Explanation:
Okay, so I need to figure out what vector a reflection takes the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to, given that the same reflection takes $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$. Hmm, reflections in linear algebra... Let me recall. A reflection is a type of linear transformation, right? So it should be represented by a matrix. But reflections in 2D can be over a line, so I need to figure out the line over which the reflection is happening.

First, let me remember that if a reflection takes vector $ \mathbf{v} $ to $ \mathbf{v}' $, then the line of reflection is the perpendicular bisector of the segment joining $ \mathbf{v} $ and $ \mathbf{v}' $. So maybe I can use that to find the line of reflection?

Let me try that approach. Let's denote the original vector as $ \mathbf{v} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} $ and its image after reflection as $ \mathbf{v}' = \begin{pmatrix} 4 \\ 3 \end{pmatrix} $. The line of reflection should be the set of points equidistant from $ \mathbf{v} $ and $ \mathbf{v}' $, which is the perpendicular bisector of the segment connecting these two points.

First, let me find the midpoint between $ \mathbf{v} $ and $ \mathbf{v}' $. The midpoint $ M $ would be:

$$
M = \frac{1}{2}(\mathbf{v} + \mathbf{v}') = \frac{1}{2}\begin{pmatrix} 5 + 4 \\ 0 + 3 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 9 \\ 3 \end{pmatrix} = \begin{pmatrix} 4.5 \\ 1.5 \end{pmatrix}
$$

Okay, so the midpoint is $ (4.5, 1.5) $. Now, the line of reflection must pass through this midpoint. Next, I need the direction of the line of reflection. Since it's the perpendicular bisector, its direction is perpendicular to the vector connecting $ \mathbf{v} $ and $ \mathbf{v}' $.

Let me compute that vector:

$$
\mathbf{v}' - \mathbf{v} = \begin{pmatrix} 4 - 5 \\ 3 - 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}
$$

So the direction vector of the segment is $ \begin{pmatrix} -1 \\ 3 \end{pmatrix} $, so the line of reflection is perpendicular to this. The perpendicular direction would have a slope that is the negative reciprocal of the slope of this vector.

Let me compute the slope of the vector $ \begin{pmatrix} -1 \\ 3 \end{pmatrix} $. The slope is $ \frac{3}{-1} = -3 $. Therefore, the slope of the line of reflection is the negative reciprocal of that, which is $ \frac{1}{3} $.

Therefore, the line of reflection has slope $ \frac{1}{3} $ and passes through the midpoint $ (4.5, 1.5) $. Let me write the equation of this line.

Using point-slope form:

$$
y - 1.5 = \frac{1}{3}(x - 4.5)
$$

Simplifying:

Multiply both sides by 3 to eliminate the denominator:

$$
3(y - 1.5) = x - 4.5 \\
3y - 4.5 = x - 4.5 \\
3y = x \\
x = 3y
$$

Wait, so the equation of the line of reflection is $ x = 3y $, or rearranged as $ y = \frac{1}{3}x $. That seems correct. Let me check with the midpoint. If $ x = 4.5 $, then $ y = 4.5 / 3 = 1.5 $, which matches the midpoint. And the slope is indeed 1/3, which is perpendicular to the original vector's slope of -3. So that seems good.

Okay, so the line of reflection is $ y = \frac{1}{3}x $. Now, I need to find the reflection of the vector $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ over this line.

Hmm, how do I reflect a point over a line in 2D? I remember there is a formula for reflecting a point over a line. Let me recall. If the line is given by $ ax + by + c = 0 $, then the reflection of a point $ (x_0, y_0) $ is given by:

$$
x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2} \\
y' = y_0 - \frac{2b(ax_0 + by_0 + c)}{a^2 + b^2}
$$

Alternatively, since our line is $ y = \frac{1}{3}x $, let me write it in standard form. $ y - \frac{1}{3}x = 0 $, so multiplying both sides by 3 to eliminate the fraction: $ -x + 3y = 0 $, so $ -x + 3y = 0 $. Therefore, $ a = -1 $, $ b = 3 $, $ c = 0 $.

So applying the formula for reflection:

For a point $ (x_0, y_0) $, the reflection $ (x', y') $ is:

$$
x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2} \\
y' = y_0 - \frac{2b(ax_0 + by_0 + c)}{a^2 + b^2}
$$

Let me compute $ ax_0 + by_0 + c $ for our point $ (-2, 3) $:

$ a = -1 $, $ b = 3 $, $ c = 0 $:

$$
(-1)(-2) + 3(3) + 0 = 2 + 9 = 11
$$

Then $ a^2 + b^2 = (-1)^2 + 3^2 = 1 + 9 = 10 $

So, compute the terms:

First, compute numerator for $ x' $:

$ - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2} = - \frac{2(-1)(11)}{10} = - \frac{-22}{10} = \frac{22}{10} = 2.2 $

Wait, hold on. Wait the formula says:

$$
x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2}
$$

Wait, is that correct? Let me check the formula again. Because I might be mixing up the formulas. Alternatively, maybe I can recall that the reflection formula for a line $ ax + by + c = 0 $ is:

$$
x' = x - \frac{2a(ax + by + c)}{a^2 + b^2} \\
y' = y - \frac{2b(ax + by + c)}{a^2 + b^2}
$$

Wait, that seems to be the case. So, if that's the case, then:

So for our case, $ a = -1 $, $ b = 3 $, $ c = 0 $, so:

Compute:

$ ax + by + c = -x + 3y + 0 = -x + 3y $

So for $ (x, y) = (-2, 3) $, that is:

$ -(-2) + 3(3) = 2 + 9 = 11 $, which matches.

Therefore:

$ x' = x - \frac{2a(-x + 3y)}{a^2 + b^2} $

Wait, but wait:

Wait, according to the formula:

$$
x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2}
$$

So plugging in:

$ x' = (-2) - \frac{2(-1)(11)}{10} = -2 - \frac{-22}{10} = -2 + 2.2 = 0.2 $

Similarly for $ y' $:

$ y' = 3 - \frac{2b(ax_0 + by_0 + c)}{a^2 + b^2} = 3 - \frac{2*3*(11)}{10} = 3 - \frac{66}{10} = 3 - 6.6 = -3.6 $

So the reflected point is $ (0.2, -3.6) $. Let me write that as fractions to be precise.

Since 0.2 is 1/5 and -3.6 is -18/5. Let me verify my calculations step by step.

First, $ x' $:

-2 - [2*(-1)*11]/10 = -2 - [ -22/10 ] = -2 + 22/10 = -2 + 11/5 = (-10/5 + 11/5) = 1/5. Yes, that's 0.2.

Similarly for y':

3 - [2*3*11]/10 = 3 - 66/10 = 3 - 6.6 = -3.6. Which is -18/5. So $ \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $.

But let me verify whether this is correct. Alternatively, maybe I can use another method to confirm.

Another way to find the reflection over the line $ y = \frac{1}{3}x $ is to use the formula for reflection over a line through the origin. Wait, our line passes through the origin? Let me check. The line is $ y = (1/3)x $, which does pass through the origin (0,0). Therefore, since the line passes through the origin, the reflection matrix can be found using some standard formula.

Alternatively, since reflections over lines through the origin can be represented by matrices. Let me recall that if we have a line making an angle $ \theta $ with the x-axis, then the reflection matrix is:

$$
\begin{pmatrix}
\cos 2\theta & \sin 2\theta \\
\sin 2\theta & -\cos 2\theta
\end{pmatrix}
$$

So maybe I can find $ \theta $ for our line. Since the line is $ y = (1/3)x $, its slope is $ m = 1/3 $. The angle $ \theta $ that the line makes with the x-axis satisfies $ \tan \theta = 1/3 $. Therefore, $ \theta = \arctan(1/3) $. Then $ 2\theta $ is the angle for the reflection matrix.

But maybe it's easier to compute the reflection matrix using another approach. Alternatively, since I know that the reflection matrix can be constructed using the formula based on the direction vector of the line.

Alternatively, let me recall that if the line is given by $ \mathbf{u} \cdot \mathbf{x} = 0 $, then the reflection matrix can be given by $ 2\mathbf{u}\mathbf{u}^T - I $, but wait, that's for a line through the origin. Wait, more accurately, if the line is spanned by a unit vector $ \mathbf{u} $, then the reflection matrix is $ 2\mathbf{u}\mathbf{u}^T - I $. Let me verify this formula.

Alternatively, maybe I can use the formula for reflecting a vector over a line with direction vector $ \mathbf{v} $. Let me recall that if we have a vector $ \mathbf{a} $, then the reflection over the line spanned by $ \mathbf{v} $ is:

$$
\text{Ref}_{\mathbf{v}}(\mathbf{a}) = 2 \frac{\mathbf{a} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} - \mathbf{a}
$$

Yes, that formula seems right. So if I can find a direction vector for the line of reflection, then I can apply this formula.

Our line of reflection is $ y = \frac{1}{3}x $, which can be represented by direction vector $ \begin{pmatrix} 3 \\ 1 \end{pmatrix} $, since the slope is 1/3. Alternatively, any scalar multiple. So taking $ \mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} $.

So let's compute the reflection of vector $ \mathbf{a} = \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ over the line spanned by $ \mathbf{v} $.

First, compute $ \mathbf{a} \cdot \mathbf{v} $:

$ (-2)(3) + (3)(1) = -6 + 3 = -3 $.

Then compute $ \mathbf{v} \cdot \mathbf{v} $:

$ 3^2 + 1^2 = 9 + 1 = 10 $.

So the formula gives:

$$
\text{Ref}_{\mathbf{v}}(\mathbf{a}) = 2 \cdot \frac{-3}{10} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}
$$

Compute each part:

First, $ 2 \cdot \frac{-3}{10} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \frac{-6}{10} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \frac{-3}{5} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -9/5 \\ -3/5 \end{pmatrix} $

Then subtract $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $, which is the same as adding $ \begin{pmatrix} 2 \\ -3 \end{pmatrix} $:

So adding:

$ \begin{pmatrix} -9/5 + 2 \\ -3/5 - 3 \end{pmatrix} = \begin{pmatrix} -9/5 + 10/5 \\ -3/5 - 15/5 \end{pmatrix} = \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $

Which matches the result I obtained earlier! So $ \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $, which is $ \begin{pmatrix} 0.2 \\ -3.6 \end{pmatrix} $. So that seems correct.

But let me verify with another approach just to be sure. Let's take the original vector $ \begin{pmatrix} 5 \\ 0 \end{pmatrix} $ and check if reflecting it over the line $ y = \frac{1}{3}x $ gives $ \begin{pmatrix} 4 \\ 3 \end{pmatrix} $.

Applying the formula again.

Let me compute reflection of $ \mathbf{a} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} $ over direction vector $ \mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} $.

First, $ \mathbf{a} \cdot \mathbf{v} = 5*3 + 0*1 = 15 $.

$ \mathbf{v} \cdot \mathbf{v} = 10 $.

Therefore:

$$
\text{Ref}_{\mathbf{v}}(\mathbf{a}) = 2*(15/10)*\begin{pmatrix} 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 5 \\ 0 \end{pmatrix} = (30/10)\begin{pmatrix} 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 5 \\ 0 \end{pmatrix} = 3*\begin{pmatrix} 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \end{pmatrix} - \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Perfect! That matches the given information. So the formula is working. Therefore, my calculation for reflecting $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ must be correct. Therefore, the answer is $ \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $, which can be written as $ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} $.

But let me check once more using the formula for reflection over a line. Let me use another formula. Let me recall that to reflect a point over a line, you can find the projection of the point onto the line, then double that projection and subtract from the original point. So:

If $ P $ is the original point, $ Q $ is its reflection, and $ M $ is the projection of $ P $ onto the line, then $ Q = 2M - P $.

So let me compute the projection of $ (-2, 3) $ onto the line $ y = (1/3)x $, then multiply by 2 and subtract the original point.

How do I find the projection of a point onto a line? Let me recall that if the line is given by $ \mathbf{r} = t \mathbf{v} $, where $ \mathbf{v} $ is a direction vector, then the projection of a point $ \mathbf{a} $ is given by $ \frac{ \mathbf{a} \cdot \mathbf{v} }{ \mathbf{v} \cdot \mathbf{v} } \mathbf{v} $.

So using $ \mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} $:

Compute $ \mathbf{a} \cdot \mathbf{v} = (-2)(3) + 3(1) = -6 + 3 = -3 $

Then $ \mathbf{v} \cdot \mathbf{v} = 10 $, so the projection vector is $ (-3/10) \mathbf{v} = (-3/10)\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -9/10 \\ -3/10 \end{pmatrix} $

Then the reflection $ Q = 2M - P $:

So $ 2M = 2*\begin{pmatrix} -9/10 \\ -3/10 \end{pmatrix} = \begin{pmatrix} -9/5 \\ -3/5 \end{pmatrix} $

Then subtract $ P = \begin{pmatrix} -2 \\ 3 \end{pmatrix} $:

So $ Q = \begin{pmatrix} -9/5 - (-2) \\ -3/5 - 3 \end{pmatrix} = \begin{pmatrix} -9/5 + 10/5 \\ -3/5 - 15/5 \end{pmatrix} = \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $

Same result again. So that's consistent. So this seems to confirm that my answer is correct.

Alternatively, maybe I can visualize or check with coordinates. Let me see.

Another way to check: Let me take the original vector $ (-2, 3) $. If I reflect it over the line $ y = (1/3)x $, which is a line passing through the origin with a gentle slope. The original point (-2, 3) is in the second quadrant. The reflection over this line should be... Hmm, perhaps it's better to compute coordinates.

But since all methods so far have given me the same answer, and the check with the given reflection of (5,0) to (4,3) works, I think my answer is correct.

But just to make sure, let me do one more check. Let me compute the reflection matrix explicitly for the line $ y = (1/3)x $, and then apply it to the vector $ (-2, 3) $.

Earlier, I mentioned that if the line is at an angle $ \theta $, then the reflection matrix is:

$$
\begin{pmatrix}
\cos 2\theta & \sin 2\theta \\
\sin 2\theta & -\cos 2\theta
\end{pmatrix}
$$

Let me compute that. Given that the line $ y = (1/3)x $ makes an angle $ \theta $ with the x-axis where $ \tan \theta = 1/3 $. So to compute $ \cos 2\theta $ and $ \sin 2\theta $, we can use double-angle formulas.

We can recall that:

$ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} $

and

$ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} $

So since $ \tan \theta = 1/3 $:

Compute $ \tan^2 \theta = 1/9 $

So:

$ \cos 2\theta = \frac{1 - 1/9}{1 + 1/9} = \frac{8/9}{10/9} = \frac{8}{10} = 4/5 $

Similarly,

$ \sin 2\theta = \frac{2*(1/3)}{1 + 1/9} = \frac{2/3}{10/9} = \frac{2/3 * 9/10} = \frac{18}{30} = 3/5 $

So the reflection matrix becomes:

$$
\begin{pmatrix}
4/5 & 3/5 \\
3/5 & -4/5
\end{pmatrix}
$$

Now, let's apply this matrix to the vector $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $:

Compute the first component:

$ (4/5)(-2) + (3/5)(3) = (-8/5) + (9/5) = 1/5 $

Second component:

$ (3/5)(-2) + (-4/5)(3) = (-6/5) + (-12/5) = (-18)/5 $

So again, we get $ \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $, which matches all previous results. Perfect! So that's the answer.

Therefore, the reflection takes $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ to $ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} $.

But just to make sure I didn't make any miscalculations in the matrix multiplication:

First component:

$ (4/5)(-2) = -8/5 $

$ (3/5)(3) = 9/5 $

Adding those together: $ -8/5 + 9/5 = 1/5 $. Correct.

Second component:

$ (3/5)(-2) = -6/5 $

$ (-4/5)(3) = -12/5 $

Adding those: $ -6/5 - 12/5 = -18/5 $. Correct.

So all methods converge to the same answer. Therefore, I'm confident that the correct answer is $ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} $.

Alternatively, maybe I can check if the midpoint between $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ and its reflection lies on the line of reflection. Let me check that.

The midpoint between $ (-2, 3) $ and $ (1/5, -18/5) $ should lie on the line $ y = (1/3)x $.

Midpoint coordinates:

x-coordinate: $ (-2 + 1/5)/2 = (-10/5 + 1/5)/2 = (-9/5)/2 = -9/10 $

y-coordinate: $ (3 + (-18/5))/2 = (15/5 - 18/5)/2 = (-3/5)/2 = -3/10 $

Now check if this midpoint lies on the line $ y = (1/3)x $:

Left-hand side (y): -3/10

Right-hand side (1/3)x: (1/3)(-9/10) = -3/10

Yes! So the midpoint lies on the line, which confirms that the reflection is correct. That's another check.

Therefore, I think I can confidently say that the answer is $ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} $.

**Final Answer**
The reflection takes $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to $\boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}$.

To solve the problem, we are told that a reflection takes the vector
$$
\begin{pmatrix} 5 \\ 0 \end{pmatrix}
$$
to
$$
\begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$
and we are asked to find where the reflection takes
$$
\begin{pmatrix} -2 \\ 3 \end{pmatrix}
$$

---

### Step 1: Understand the Nature of the Reflection

Since the reflection is a **linear transformation**, it must preserve the origin (if the reflection line passes through the origin), and it corresponds to a **reflection matrix**, which is a 2×2 orthogonal matrix. The reflection will take any point on the line to itself, and any point off the line to its mirror image with respect to that line.

---

### Step 2: Determine the Line of Reflection

To find the line of reflection, we use the fact that the **midpoint** of the original and reflected points lies on the line, and that the **line is perpendicular** to the vector joining the two points.

Let’s denote:
- $ \vec{v} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} $
- $ \vec{v}' = \begin{pmatrix} 4 \\ 3 \end{pmatrix} $

Then, the **midpoint** is:
$$
\text{Midpoint} = \frac{1}{2}(\vec{v} + \vec{v}') = \frac{1}{2} \begin{pmatrix} 9 \\ 3 \end{pmatrix} = \begin{pmatrix} 4.5 \\ 1.5 \end{pmatrix}
$$

The vector connecting the two points is:
$$
\vec{v}' - \vec{v} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}
$$

This vector has slope $-\frac{3}{1} = -3$, so the **line of reflection** (which is perpendicular to this vector) has a slope of $ \frac{1}{3} $. Since the midpoint lies on this line, the line of reflection is a straight line through that point with slope $ \frac{1}{3} $, so its equation is:
$$
y = \frac{1}{3}x
$$

---

### Step 3: Reflect the Point $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ Over the Line

We now want to reflect the point $ (-2, 3) $ over the line $ y = \frac{1}{3}x $.

#### Method: Use the Reflection Matrix

Reflection over a line that passes through the origin can be done using this standard matrix:
$$
R = \begin{pmatrix}
\cos(2\theta) & \sin(2\theta) \\
\sin(2\theta) & -\cos(2\theta)
\end{pmatrix}
$$
where $ \theta $ is the angle of the line with the x-axis.

Since the line $ y = \frac{1}{3}x $ has slope $ \frac{1}{3} $, we compute:
- $ \tan(\theta) = \frac{1}{3} $
- $ \sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)} = \frac{2/3}{10/9} = \frac{3}{5} $
- $ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \frac{8/9}{10/9} = \frac{4}{5} $

So the reflection matrix is:
$$
R = \begin{pmatrix}
\frac{4}{5} & \frac{3}{5} \\
\frac{3}{5} & -\frac{4}{5}
\end{pmatrix}
$$

Now apply this matrix to the vector $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $:

$$
R \cdot \begin{pmatrix} -2 \\ 3 \end{pmatrix}
=
\begin{pmatrix}
\frac{4}{5} \cdot (-2) + \frac{3}{5} \cdot 3 \\
\frac{3}{5} \cdot (-2) + \left(-\frac{4}{5}\right) \cdot 3
\end{pmatrix}
=
\begin{pmatrix}
\frac{-8 + 9}{5} \\
\frac{-6 - 12}{5}
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{5} \\
-\frac{18}{5}
\end{pmatrix}
$$

---

### Final Answer

$$
\boxed{ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} }
$$

[Assistant]
Explanation:
Okay, so I need to figure out what vector a reflection takes the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to, given that the same reflection takes $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$. Hmm, reflections in linear algebra... Let me recall. A reflection is a type of linear transformation, right? So it should be represented by a matrix. But reflections in 2D can be over a line, so I need to figure out the line over which the reflection is happening.

First, let me remember that if a reflection takes vector $ \mathbf{v} $ to $ \mathbf{v}' $, then the line of reflection is the perpendicular bisector of the segment joining $ \mathbf{v} $ and $ \mathbf{v}' $. So maybe I can use that to find the line of reflection?

Let me try that approach. Let's denote the original vector as $ \mathbf{v} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} $ and its image after reflection as $ \mathbf{v}' = \begin{pmatrix} 4 \\ 3 \end{pmatrix} $. The line of reflection should be the set of points equidistant from $ \mathbf{v} $ and $ \mathbf{v}' $, which is the perpendicular bisector of the segment connecting these two points.

First, let me find the midpoint between $ \mathbf{v} $ and $ \mathbf{v}' $. The midpoint $ M $ would be:

$$
M = \frac{1}{2}(\mathbf{v} + \mathbf{v}') = \frac{1}{2}\begin{pmatrix} 5 + 4 \\ 0 + 3 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 9 \\ 3 \end{pmatrix} = \begin{pmatrix} 4.5 \\ 1.5 \end{pmatrix}
$$

Okay, so the midpoint is $ (4.5, 1.5) $. Now, the line of reflection must pass through this midpoint. Next, I need the direction of the line of reflection. Since it's the perpendicular bisector, its direction is perpendicular to the vector connecting $ \mathbf{v} $ and $ \mathbf{v}' $.

Let me compute that vector:

$$
\mathbf{v}' - \mathbf{v} = \begin{pmatrix} 4 - 5 \\ 3 - 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}
$$

So the direction vector of the segment is $ \begin{pmatrix} -1 \\ 3 \end{pmatrix} $, so the line of reflection is perpendicular to this. The perpendicular direction would have a slope that is the negative reciprocal of the slope of this vector.

Let me compute the slope of the vector $ \begin{pmatrix} -1 \\ 3 \end{pmatrix} $. The slope is $ \frac{3}{-1} = -3 $. Therefore, the slope of the line of reflection is the negative reciprocal of that, which is $ \frac{1}{3} $.

Therefore, the line of reflection has slope $ \frac{1}{3} $ and passes through the midpoint $ (4.5, 1.5) $. Let me write the equation of this line.

Using point-slope form:

$$
y - 1.5 = \frac{1}{3}(x - 4.5)
$$

Simplifying:

Multiply both sides by 3 to eliminate the denominator:

$$
3(y - 1.5) = x - 4.5 \\
3y - 4.5 = x - 4.5 \\
3y = x \\
x = 3y
$$

Wait, so the equation of the line of reflection is $ x = 3y $, or rearranged as $ y = \frac{1}{3}x $. That seems correct. Let me check with the midpoint. If $ x = 4.5 $, then $ y = 4.5 / 3 = 1.5 $, which matches the midpoint. And the slope is indeed 1/3, which is perpendicular to the original vector's slope of -3. So that seems good.

Okay, so the line of reflection is $ y = \frac{1}{3}x $. Now, I need to find the reflection of the vector $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ over this line.

Hmm, how do I reflect a point over a line in 2D? I remember there is a formula for reflecting a point over a line. Let me recall. If the line is given by $ ax + by + c = 0 $, then the reflection of a point $ (x_0, y_0) $ is given by:

$$
x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2} \\
y' = y_0 - \frac{2b(ax_0 + by_0 + c)}{a^2 + b^2}
$$

Alternatively, since our line is $ y = \frac{1}{3}x $, let me write it in standard form. $ y - \frac{1}{3}x = 0 $, so multiplying both sides by 3 to eliminate the fraction: $ -x + 3y = 0 $, so $ -x + 3y = 0 $. Therefore, $ a = -1 $, $ b = 3 $, $ c = 0 $.

So applying the formula for reflection:

For a point $ (x_0, y_0) $, the reflection $ (x', y') $ is:

$$
x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2} \\
y' = y_0 - \frac{2b(ax_0 + by_0 + c)}{a^2 + b^2}
$$

Let me compute $ ax_0 + by_0 + c $ for our point $ (-2, 3) $:

$ a = -1 $, $ b = 3 $, $ c = 0 $:

$$
(-1)(-2) + 3(3) + 0 = 2 + 9 = 11
$$

Then $ a^2 + b^2 = (-1)^2 + 3^2 = 1 + 9 = 10 $

So, compute the terms:

First, compute numerator for $ x' $:

$ - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2} = - \frac{2(-1)(11)}{10} = - \frac{-22}{10} = \frac{22}{10} = 2.2 $

Wait, hold on. Wait the formula says:

$$
x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2}
$$

Wait, is that correct? Let me check the formula again. Because I might be mixing up the formulas. Alternatively, maybe I can recall that the reflection formula for a line $ ax + by + c = 0 $ is:

$$
x' = x - \frac{2a(ax + by + c)}{a^2 + b^2} \\
y' = y - \frac{2b(ax + by + c)}{a^2 + b^2}
$$

Wait, that seems to be the case. So, if that's the case, then:

So for our case, $ a = -1 $, $ b = 3 $, $ c = 0 $, so:

Compute:

$ ax + by + c = -x + 3y + 0 = -x + 3y $

So for $ (x, y) = (-2, 3) $, that is:

$ -(-2) + 3(3) = 2 + 9 = 11 $, which matches.

Therefore:

$ x' = x - \frac{2a(-x + 3y)}{a^2 + b^2} $

Wait, but wait:

Wait, according to the formula:

$$
x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2}
$$

So plugging in:

$ x' = (-2) - \frac{2(-1)(11)}{10} = -2 - \frac{-22}{10} = -2 + 2.2 = 0.2 $

Similarly for $ y' $:

$ y' = 3 - \frac{2b(ax_0 + by_0 + c)}{a^2 + b^2} = 3 - \frac{2*3*(11)}{10} = 3 - \frac{66}{10} = 3 - 6.6 = -3.6 $

So the reflected point is $ (0.2, -3.6) $. Let me write that as fractions to be precise.

Since 0.2 is 1/5 and -3.6 is -18/5. Let me verify my calculations step by step.

First, $ x' $:

-2 - [2*(-1)*11]/10 = -2 - [ -22/10 ] = -2 + 22/10 = -2 + 11/5 = (-10/5 + 11/5) = 1/5. Yes, that's 0.2.

Similarly for y':

3 - [2*3*11]/10 = 3 - 66/10 = 3 - 6.6 = -3.6. Which is -18/5. So $ \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $.

But let me verify whether this is correct. Alternatively, maybe I can use another method to confirm.

Another way to find the reflection over the line $ y = \frac{1}{3}x $ is to use the formula for reflection over a line through the origin. Wait, our line passes through the origin? Let me check. The line is $ y = (1/3)x $, which does pass through the origin (0,0). Therefore, since the line passes through the origin, the reflection matrix can be found using some standard formula.

Alternatively, since reflections over lines through the origin can be represented by matrices. Let me recall that if we have a line making an angle $ \theta $ with the x-axis, then the reflection matrix is:

$$
\begin{pmatrix}
\cos 2\theta & \sin 2\theta \\
\sin 2\theta & -\cos 2\theta
\end{pmatrix}
$$

So maybe I can find $ \theta $ for our line. Since the line is $ y = (1/3)x $, its slope is $ m = 1/3 $. The angle $ \theta $ that the line makes with the x-axis satisfies $ \tan \theta = 1/3 $. Therefore, $ \theta = \arctan(1/3) $. Then $ 2\theta $ is the angle for the reflection matrix.

But maybe it's easier to compute the reflection matrix using another approach. Alternatively, since I know that the reflection matrix can be constructed using the formula based on the direction vector of the line.

Alternatively, let me recall that if the line is given by $ \mathbf{u} \cdot \mathbf{x} = 0 $, then the reflection matrix can be given by $ 2\mathbf{u}\mathbf{u}^T - I $, but wait, that's for a line through the origin. Wait, more accurately, if the line is spanned by a unit vector $ \mathbf{u} $, then the reflection matrix is $ 2\mathbf{u}\mathbf{u}^T - I $. Let me verify this formula.

Alternatively, maybe I can use the formula for reflecting a vector over a line with direction vector $ \mathbf{v} $. Let me recall that if we have a vector $ \mathbf{a} $, then the reflection over the line spanned by $ \mathbf{v} $ is:

$$
\text{Ref}_{\mathbf{v}}(\mathbf{a}) = 2 \frac{\mathbf{a} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} - \mathbf{a}
$$

Yes, that formula seems right. So if I can find a direction vector for the line of reflection, then I can apply this formula.

Our line of reflection is $ y = \frac{1}{3}x $, which can be represented by direction vector $ \begin{pmatrix} 3 \\ 1 \end{pmatrix} $, since the slope is 1/3. Alternatively, any scalar multiple. So taking $ \mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} $.

So let's compute the reflection of vector $ \mathbf{a} = \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ over the line spanned by $ \mathbf{v} $.

First, compute $ \mathbf{a} \cdot \mathbf{v} $:

$ (-2)(3) + (3)(1) = -6 + 3 = -3 $.

Then compute $ \mathbf{v} \cdot \mathbf{v} $:

$ 3^2 + 1^2 = 9 + 1 = 10 $.

So the formula gives:

$$
\text{Ref}_{\mathbf{v}}(\mathbf{a}) = 2 \cdot \frac{-3}{10} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}
$$

Compute each part:

First, $ 2 \cdot \frac{-3}{10} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \frac{-6}{10} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \frac{-3}{5} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -9/5 \\ -3/5 \end{pmatrix} $

Then subtract $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $, which is the same as adding $ \begin{pmatrix} 2 \\ -3 \end{pmatrix} $:

So adding:

$ \begin{pmatrix} -9/5 + 2 \\ -3/5 - 3 \end{pmatrix} = \begin{pmatrix} -9/5 + 10/5 \\ -3/5 - 15/5 \end{pmatrix} = \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $

Which matches the result I obtained earlier! So $ \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $, which is $ \begin{pmatrix} 0.2 \\ -3.6 \end{pmatrix} $. So that seems correct.

But let me verify with another approach just to be sure. Let's take the original vector $ \begin{pmatrix} 5 \\ 0 \end{pmatrix} $ and check if reflecting it over the line $ y = \frac{1}{3}x $ gives $ \begin{pmatrix} 4 \\ 3 \end{pmatrix} $.

Applying the formula again.

Let me compute reflection of $ \mathbf{a} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} $ over direction vector $ \mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} $.

First, $ \mathbf{a} \cdot \mathbf{v} = 5*3 + 0*1 = 15 $.

$ \mathbf{v} \cdot \mathbf{v} = 10 $.

Therefore:

$$
\text{Ref}_{\mathbf{v}}(\mathbf{a}) = 2*(15/10)*\begin{pmatrix} 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 5 \\ 0 \end{pmatrix} = (30/10)\begin{pmatrix} 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 5 \\ 0 \end{pmatrix} = 3*\begin{pmatrix} 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 9 \\ 3 \end{pmatrix} - \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Perfect! That matches the given information. So the formula is working. Therefore, my calculation for reflecting $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ must be correct. Therefore, the answer is $ \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $, which can be written as $ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} $.

But let me check once more using the formula for reflection over a line. Let me use another formula. Let me recall that to reflect a point over a line, you can find the projection of the point onto the line, then double that projection and subtract from the original point. So:

If $ P $ is the original point, $ Q $ is its reflection, and $ M $ is the projection of $ P $ onto the line, then $ Q = 2M - P $.

So let me compute the projection of $ (-2, 3) $ onto the line $ y = (1/3)x $, then multiply by 2 and subtract the original point.

How do I find the projection of a point onto a line? Let me recall that if the line is given by $ \mathbf{r} = t \mathbf{v} $, where $ \mathbf{v} $ is a direction vector, then the projection of a point $ \mathbf{a} $ is given by $ \frac{ \mathbf{a} \cdot \mathbf{v} }{ \mathbf{v} \cdot \mathbf{v} } \mathbf{v} $.

So using $ \mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} $:

Compute $ \mathbf{a} \cdot \mathbf{v} = (-2)(3) + 3(1) = -6 + 3 = -3 $

Then $ \mathbf{v} \cdot \mathbf{v} = 10 $, so the projection vector is $ (-3/10) \mathbf{v} = (-3/10)\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -9/10 \\ -3/10 \end{pmatrix} $

Then the reflection $ Q = 2M - P $:

So $ 2M = 2*\begin{pmatrix} -9/10 \\ -3/10 \end{pmatrix} = \begin{pmatrix} -9/5 \\ -3/5 \end{pmatrix} $

Then subtract $ P = \begin{pmatrix} -2 \\ 3 \end{pmatrix} $:

So $ Q = \begin{pmatrix} -9/5 - (-2) \\ -3/5 - 3 \end{pmatrix} = \begin{pmatrix} -9/5 + 10/5 \\ -3/5 - 15/5 \end{pmatrix} = \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $

Same result again. So that's consistent. So this seems to confirm that my answer is correct.

Alternatively, maybe I can visualize or check with coordinates. Let me see.

Another way to check: Let me take the original vector $ (-2, 3) $. If I reflect it over the line $ y = (1/3)x $, which is a line passing through the origin with a gentle slope. The original point (-2, 3) is in the second quadrant. The reflection over this line should be... Hmm, perhaps it's better to compute coordinates.

But since all methods so far have given me the same answer, and the check with the given reflection of (5,0) to (4,3) works, I think my answer is correct.

But just to make sure, let me do one more check. Let me compute the reflection matrix explicitly for the line $ y = (1/3)x $, and then apply it to the vector $ (-2, 3) $.

Earlier, I mentioned that if the line is at an angle $ \theta $, then the reflection matrix is:

$$
\begin{pmatrix}
\cos 2\theta & \sin 2\theta \\
\sin 2\theta & -\cos 2\theta
\end{pmatrix}
$$

Let me compute that. Given that the line $ y = (1/3)x $ makes an angle $ \theta $ with the x-axis where $ \tan \theta = 1/3 $. So to compute $ \cos 2\theta $ and $ \sin 2\theta $, we can use double-angle formulas.

We can recall that:

$ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} $

and

$ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} $

So since $ \tan \theta = 1/3 $:

Compute $ \tan^2 \theta = 1/9 $

So:

$ \cos 2\theta = \frac{1 - 1/9}{1 + 1/9} = \frac{8/9}{10/9} = \frac{8}{10} = 4/5 $

Similarly,

$ \sin 2\theta = \frac{2*(1/3)}{1 + 1/9} = \frac{2/3}{10/9} = \frac{2/3 * 9/10} = \frac{18}{30} = 3/5 $

So the reflection matrix becomes:

$$
\begin{pmatrix}
4/5 & 3/5 \\
3/5 & -4/5
\end{pmatrix}
$$

Now, let's apply this matrix to the vector $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $:

Compute the first component:

$ (4/5)(-2) + (3/5)(3) = (-8/5) + (9/5) = 1/5 $

Second component:

$ (3/5)(-2) + (-4/5)(3) = (-6/5) + (-12/5) = (-18)/5 $

So again, we get $ \begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix} $, which matches all previous results. Perfect! So that's the answer.

Therefore, the reflection takes $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ to $ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} $.

But just to make sure I didn't make any miscalculations in the matrix multiplication:

First component:

$ (4/5)(-2) = -8/5 $

$ (3/5)(3) = 9/5 $

Adding those together: $ -8/5 + 9/5 = 1/5 $. Correct.

Second component:

$ (3/5)(-2) = -6/5 $

$ (-4/5)(3) = -12/5 $

Adding those: $ -6/5 - 12/5 = -18/5 $. Correct.

So all methods converge to the same answer. Therefore, I'm confident that the correct answer is $ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} $.

Alternatively, maybe I can check if the midpoint between $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $ and its reflection lies on the line of reflection. Let me check that.

The midpoint between $ (-2, 3) $ and $ (1/5, -18/5) $ should lie on the line $ y = (1/3)x $.

Midpoint coordinates:

x-coordinate: $ (-2 + 1/5)/2 = (-10/5 + 1/5)/2 = (-9/5)/2 = -9/10 $

y-coordinate: $ (3 + (-18/5))/2 = (15/5 - 18/5)/2 = (-3/5)/2 = -3/10 $

Now check if this midpoint lies on the line $ y = (1/3)x $:

Left-hand side (y): -3/10

Right-hand side (1/3)x: (1/3)(-9/10) = -3/10

Yes! So the midpoint lies on the line, which confirms that the reflection is correct. That's another check.

Therefore, I think I can confidently say that the answer is $ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} $.

**Final Answer**
The reflection takes $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to $\boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}$.

To solve the problem, we are told that a reflection takes the vector
$$
\begin{pmatrix} 5 \\ 0 \end{pmatrix}
$$
to
$$
\begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$
and we are asked to find where the reflection takes
$$
\begin{pmatrix} -2 \\ 3 \end{pmatrix}
$$

---

### Step 1: Understand the Nature of the Reflection

Since the reflection is a **linear transformation**, it must preserve the origin (if the reflection line passes through the origin), and it corresponds to a **reflection matrix**, which is a 2×2 orthogonal matrix. The reflection will take any point on the line to itself, and any point off the line to its mirror image with respect to that line.

---

### Step 2: Determine the Line of Reflection

To find the line of reflection, we use the fact that the **midpoint** of the original and reflected points lies on the line, and that the **line is perpendicular** to the vector joining the two points.

Let’s denote:
- $ \vec{v} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} $
- $ \vec{v}' = \begin{pmatrix} 4 \\ 3 \end{pmatrix} $

Then, the **midpoint** is:
$$
\text{Midpoint} = \frac{1}{2}(\vec{v} + \vec{v}') = \frac{1}{2} \begin{pmatrix} 9 \\ 3 \end{pmatrix} = \begin{pmatrix} 4.5 \\ 1.5 \end{pmatrix}
$$

The vector connecting the two points is:
$$
\vec{v}' - \vec{v} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}
$$

This vector has slope $-\frac{3}{1} = -3$, so the **line of reflection** (which is perpendicular to this vector) has a slope of $ \frac{1}{3} $. Since the midpoint lies on this line, the line of reflection is a straight line through that point with slope $ \frac{1}{3} $, so its equation is:
$$
y = \frac{1}{3}x
$$

---

### Step 3: Reflect the Point $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ Over the Line

We now want to reflect the point $ (-2, 3) $ over the line $ y = \frac{1}{3}x $.

#### Method: Use the Reflection Matrix

Reflection over a line that passes through the origin can be done using this standard matrix:
$$
R = \begin{pmatrix}
\cos(2\theta) & \sin(2\theta) \\
\sin(2\theta) & -\cos(2\theta)
\end{pmatrix}
$$
where $ \theta $ is the angle of the line with the x-axis.

Since the line $ y = \frac{1}{3}x $ has slope $ \frac{1}{3} $, we compute:
- $ \tan(\theta) = \frac{1}{3} $
- $ \sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)} = \frac{2/3}{10/9} = \frac{3}{5} $
- $ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \frac{8/9}{10/9} = \frac{4}{5} $

So the reflection matrix is:
$$
R = \begin{pmatrix}
\frac{4}{5} & \frac{3}{5} \\
\frac{3}{5} & -\frac{4}{5}
\end{pmatrix}
$$

Now apply this matrix to the vector $ \begin{pmatrix} -2 \\ 3 \end{pmatrix} $:

$$
R \cdot \begin{pmatrix} -2 \\ 3 \end{pmatrix}
=
\begin{pmatrix}
\frac{4}{5} \cdot (-2) + \frac{3}{5} \cdot 3 \\
\frac{3}{5} \cdot (-2) + \left(-\frac{4}{5}\right) \cdot 3
\end{pmatrix}
=
\begin{pmatrix}
\frac{-8 + 9}{5} \\
\frac{-6 - 12}{5}
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{5} \\
-\frac{18}{5}
\end{pmatrix}
$$

---

### Final Answer

$$
\boxed{ \begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix} }
$$

Solution:
The midpoint of $(5,0)$ and $(4,3)$ is
\[\left( \frac{5 + 4}{2}, \frac{0 + 3}{2} \right) = \left( \frac{9}{2}, \frac{3}{2} \right).\]This tells us that the vector being reflected over is a scalar multiple of $\begin{pmatrix} \frac{9}{2} \\ \frac{3}{2} \end{pmatrix}.$  We can then assume that the vector being reflected over is $\begin{pmatrix} 3 \\ 1 \end{pmatrix}.$

[asy]
usepackage("amsmath");

unitsize(1 cm);

pair A, B, M, O, R, S;

O = (0,0);
A = (5,0);
R = (4,3);
B = (-2,3);
S = (1/5,-18/5);
M = (A + R)/2;

draw((-3,-1)--(5,5/3),red + dashed);
draw(O--M,red,Arrow(6));
draw((-4,0)--(6,0));
draw((0,-4)--(0,4));
draw(O--A,Arrow(6));
draw(O--R,Arrow(6));
draw(A--R,dashed,Arrow(6));
draw(O--B,Arrow(6));
draw(O--S,Arrow(6));
draw(B--S,dashed,Arrow(6));
label("$\begin{pmatrix} 5 \\ 0 \end{pmatrix}$", A, S);
label("$\begin{pmatrix} 4 \\ 3 \end{pmatrix}$", R, NE);
label("$\begin{pmatrix} -2 \\ 3 \end{pmatrix}$", B, NW);
label("$\begin{pmatrix} \frac{9}{2} \\ \frac{3}{2} \end{pmatrix}$", M, N);
[/asy]

The projection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ onto $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ is
\[\operatorname{proj}_{\begin{pmatrix} 3 \\ 1 \end{pmatrix}} \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \frac{\begin{pmatrix} -2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix}}{\begin{pmatrix} 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix}} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \frac{-3}{10} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -\frac{9}{10} \\ -\frac{3}{10} \end{pmatrix}.\]Hence, the reflection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ is $2 \begin{pmatrix} -\frac{9}{10} \\ -\frac{3}{10} \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \boxed{\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}}.$


Explanation:
Okay, so I need to find the remainder when $129^{34} + 96^{38}$ is divided by 11. Hmm, let me think about how to approach this. I remember something about modular arithmetic, maybe Fermat's Little Theorem? Let me recall...

Fermat's Little Theorem says that if p is a prime number, then for any integer a not divisible by p, $a^{p-1} \equiv 1 \mod p$. Since 11 is a prime number, this theorem might be useful here. Let me check if 129 and 96 are divisible by 11.

First, let me find 129 mod 11. Dividing 129 by 11... 11*11 is 121, so 129 - 121 is 8. So 129 ≡ 8 mod 11. Similarly, 96 divided by 11... 11*8 is 88, so 96 - 88 is 8. Wait, so 96 is also congruent to 8 mod 11? That's interesting. So both 129 and 96 leave a remainder of 8 when divided by 11. So maybe I can rewrite the original expression as $8^{34} + 8^{38}$ mod 11?

Yes, that simplifies things. So now I need to compute $8^{34} + 8^{38} \mod 11$. Let me see. Maybe I can factor out some common terms. Let's see, both terms have 8^{34}, so $8^{34}(1 + 8^{4})$ mod 11. Let me check that exponent. 38 - 34 is 4, right? So that's correct. So then I need to compute $8^{34} \times (1 + 8^4)$ mod 11.

Alternatively, maybe compute each term separately. Let me try that. First, let's compute 8^{34} mod 11 and 8^{38} mod 11, then add them together and take mod 11 again.

But maybe using Fermat's Little Theorem would help here. Since 11 is prime, and 8 is not divisible by 11, so according to Fermat's Little Theorem, 8^{10} ≡ 1 mod 11. So exponents can be reduced modulo 10. Let me verify that.

Yes, because 8^{10} ≡ 1 mod 11, so 8^{k} ≡ 8^{k mod 10} mod 11. Wait, not exactly. It's 8^{10} ≡ 1, so 8^{10m + r} ≡ (8^{10})^m * 8^r ≡ 1^m * 8^r ≡ 8^r mod 11. Therefore, the exponent can be reduced modulo 10.

Therefore, let me apply this. For 8^{34}, divide 34 by 10. 34 divided by 10 is 3 with a remainder of 4. So 8^{34} ≡ 8^4 mod 11. Similarly, for 8^{38}, divide 38 by 10. 38 divided by 10 is 3 with remainder 8, so 8^{38} ≡ 8^8 mod 11.

Therefore, the original expression becomes 8^4 + 8^8 mod 11. Let me compute these powers of 8 modulo 11.

First, compute 8^1 mod 11 = 8.

8^2 = 64 mod 11. Let me divide 64 by 11. 11*5=55, so 64-55=9. So 8^2 ≡ 9 mod 11.

Then 8^3 = 8^2 * 8 ≡ 9 * 8 = 72 mod 11. 72 divided by 11 is 6*11=66, remainder 6. So 8^3 ≡ 6 mod 11.

8^4 = 8^3 * 8 ≡ 6 * 8 = 48 mod 11. 11*4=44, so 48-44=4. So 8^4 ≡ 4 mod 11.

Then 8^5 = 8^4 * 8 ≡ 4 * 8 = 32 mod 11. 11*2=22, 32-22=10. So 8^5 ≡ 10 mod 11.

8^6 = 8^5 * 8 ≡ 10 * 8 = 80 mod 11. 11*7=77, 80-77=3. So 8^6 ≡ 3 mod 11.

8^7 = 3 * 8 = 24 mod 11. 24 - 22 = 2. So 8^7 ≡ 2 mod 11.

8^8 = 2 * 8 = 16 mod 11. 16 - 11 = 5. So 8^8 ≡ 5 mod 11.

So putting it back, 8^4 + 8^8 ≡ 4 + 5 = 9 mod 11. Therefore, the remainder is 9?

Wait, but let me check again. Let me make sure I didn't make a mistake in calculating the exponents. Let me re-verify each step.

First, 8^1 = 8 mod 11. 8^2: 8*8=64, 64 mod 11: 11*5=55, 64-55=9. Correct. 8^3: 9*8=72, 72 mod 11: 11*6=66, 72-66=6. Correct. 8^4: 6*8=48, mod 11: 48-44=4. Correct. 8^5: 4*8=32, 32-22=10. Correct. 8^6: 10*8=80. 80 - 77=3. Correct. 8^7: 3*8=24. 24-22=2. Correct. 8^8: 2*8=16. 16-11=5. Correct.

So 8^4 is 4 and 8^8 is 5. So 4 + 5 = 9. Therefore, 9 mod 11. So the remainder is 9?

Wait, but let me check another way. Maybe I can also compute 8^10 mod 11, which should be 1, to check my earlier steps. Let me compute 8^10.

Alternatively, since we know that 8^5 is 10 mod 11, then 8^5 * 8^5 = 8^10 ≡ 10*10 = 100 mod 11. 100 divided by 11 is 9*11=99, so 100 -99=1. Yep, so 8^10 ≡ 1 mod 11. Which confirms Fermat's Little Theorem.

Therefore, my calculation of exponents seems correct. So 8^34 is equivalent to 8^(3*10 +4)= (8^10)^3 *8^4 ≡ 1^3 *8^4 ≡8^4 mod 11, and similarly 8^38 is 8^(3*10 +8)= (8^10)^3 *8^8 ≡ 1^3 *8^8 mod 11. So adding those gives 8^4 +8^8 ≡ 4 +5=9 mod11.

But let me check with another approach. Maybe compute 8^4 and 8^8 mod 11 using another method.

Alternatively, maybe compute 8^4 mod 11. 8^2=64≡9 mod11, then 8^4=(8^2)^2=9^2=81≡81-77=4 mod11. Correct.

Similarly 8^8=(8^4)^2=4^2=16≡5 mod11. Correct. So 4 +5=9.

Therefore, the remainder when 129^34 +96^38 is divided by 11 is 9?

But wait, let me check once more. Maybe I made a mistake in reducing the exponents? Let me confirm the exponents again.

Original exponents: 34 and 38.

For 8^34: 34 divided by 10 is 3 with remainder 4. So exponent 34 ≡4 mod10. Therefore 8^34≡8^4.

Similarly, 38 divided by 10 is 3 with remainder 8, so exponent 38≡8 mod10. Therefore 8^38≡8^8.

So that's correct.

Alternatively, perhaps I can compute 8^4 +8^8 mod11. Which is 4 +5=9.

So 9. But wait, just to be thorough, let me check with another method. Maybe compute 129^34 mod11 and 96^38 mod11 separately, using the original numbers.

Wait, we already found that 129≡8 mod11 and 96≡8 mod11, so both terms are 8^something mod11, so that's consistent.

Alternatively, maybe I can use another theorem or approach. Let me think.

Alternatively, maybe compute 8^4 + 8^8 mod11 as follows. Let me note that 8^4 is 4, 8^8 is 5. But maybe there is a pattern here. Alternatively, perhaps compute 8^4 +8^8 = 8^4(1 + 8^4). Wait, 8^4 is 4, so 1 + 8^4 is 1 +4=5. Therefore, 8^4 *5. Which is 4*5=20≡20-11=9 mod11. Same result.

So that's another way to look at it, and still gives 9. So seems like the answer is 9.

But let me check once again. Maybe I made a mistake in the calculation of 8^4 or 8^8. Let me recalculate:

Starting with 8^1 = 8 mod11.

8^2: 8*8=64. 64 divided by 11 is 5*11=55, remainder 9. So 8^2 ≡9.

8^3: 8^2 *8=9*8=72. 72 divided by 11 is 6*11=66, remainder 6. So 8^3≡6.

8^4: 8^3 *8=6*8=48. 48 divided by 11 is 4*11=44, remainder 4. So 8^4≡4.

8^5: 4*8=32. 32-2*11=32-22=10. So 8^5≡10.

8^6: 10*8=80. 80-7*11=80-77=3. So 8^6≡3.

8^7: 3*8=24. 24-2*11=2. So 8^7≡2.

8^8: 2*8=16. 16-11=5. So 8^8≡5.

Yes, same results. So adding 4 +5=9.

Therefore, I think the answer is 9. But just to be 100% sure, let me check if there is another possible mistake.

Wait, maybe the initial step where I converted 129 and 96 to mod11? Let me check again.

129 divided by 11: 11*11=121, so 129 -121=8. So 129≡8 mod11. Correct.

96 divided by 11: 11*8=88, 96-88=8. So 96≡8 mod11. Correct.

Therefore, both terms are 8^something mod11, so that's correct.

Alternatively, perhaps the problem is that 8^something is being calculated. Wait, but I did that. So seems like all steps are correct.

So the answer is 9? Let me think again. Wait, another way to check is to compute 129^34 +96^38 mod11. Since 129 ≡8 and 96≡8, then the expression is 8^34 +8^38 mod11. Which is equal to 8^34(1 +8^4) mod11, as I did before. Which is 8^34*(1 + 4)=8^34*5 mod11.

But 8^34: since 8^10≡1 mod11, then 8^34=8^(3*10 +4)= (8^10)^3 *8^4≡1^3 *8^4≡8^4≡4 mod11. Therefore, 4*5=20≡9 mod11. So same answer.

Another way. So even if I factor it differently, I still get 9.

Alternatively, maybe compute 8^34 +8^38 mod11 as follows:

Note that 8^34 +8^38 = 8^34 + 8^34 *8^4 = 8^34 (1 + 8^4) mod11. As above. Which is 8^34 * 5.

But 8^34 is 8^(3*10 +4)= (8^10)^3 *8^4 ≡1^3 *4=4 mod11. So 4*5=20≡9.

Same answer. So, all approaches give me 9.

Alternatively, maybe I can compute 8^4 +8^8 mod11 as follows:

We know that 8^4=4 and 8^8= (8^4)^2=4^2=16≡5 mod11. Therefore, same as before.

Therefore, 4+5=9.

Therefore, I think the answer is 9.

But let me check if 9 is correct by perhaps testing with smaller exponents. For example, let's take smaller exponents and see if the method works.

For example, let me compute 8^1 +8^1 = 16 mod11=5. If I apply my method: 8^1 +8^1=8^1(1 +8^0). Wait, but in that case, exponents would be different. Maybe not the best example.

Alternatively, take 8^4 +8^8. Which we calculated as 4 +5=9. If I compute 8^4 +8^8 directly:

8^4=4096. 4096 mod11. Let me compute that.

Alternatively, since we know 8^4≡4 mod11, so 4096 mod11 is 4. Similarly, 8^8= (8^4)^2= 4^2=16≡5 mod11. So adding them gives 4+5=9.

Alternatively, compute 8^4 +8^8 directly. Let me compute 8^4: 8*8=64, 64*8=512, 512*8=4096. 4096 divided by 11. Well, 11*372=4092. 4096 -4092=4. So 4096≡4 mod11.

Similarly, 8^8: 8^4=4096 ≡4, so 8^8= (8^4)^2= 4^2=16≡5 mod11. Therefore, adding gives 4 +5=9.

Yes, that's correct.

Therefore, I think the answer is 9.

Wait, but just to make sure, let me check with another approach. Maybe using negative numbers? Sometimes it's easier to work with negative residues. For example, 8 mod11 is 8, but maybe 8 is equivalent to -3 mod11 because 8 +3=11. Maybe working with -3 can simplify calculations. Let me try that.

So 8 ≡ -3 mod11. Then 8^34 +8^38 ≡ (-3)^34 + (-3)^38 mod11.

Note that (-3)^even = 3^even, since negative even power becomes positive.

So (-3)^34 = 3^34 and (-3)^38 = 3^38. Therefore, the expression is 3^34 +3^38 mod11.

But maybe this is more complicated? Let me see.

But maybe we can compute exponents of 3 mod11? Let me see.

Using Fermat's theorem again. Since 3^10 ≡1 mod11. Therefore, exponents can be reduced modulo 10.

So 3^34: 34 divided by 10 is 3*10=30, remainder 4. So 3^34 ≡3^4 mod11. Similarly, 3^38 ≡3^8 mod11.

Therefore, compute 3^4 +3^8 mod11.

Compute 3^1=3 mod11.

3^2=9 mod11.

3^3=27≡27-22=5 mod11.

3^4=5*3=15≡15-11=4 mod11.

3^5=4*3=12≡1 mod11.

3^6=1*3=3 mod11.

3^7=3*3=9 mod11.

3^8=9*3=27≡5 mod11.

Therefore, 3^4 +3^8 ≡4 +5=9 mod11. Same result!

So again, we get 9.

So even when converting 8 to -3 mod11, we still get the same answer.

Therefore, this is a good check.

Another check: compute 3^4 +3^8 mod11=4 +5=9.

Therefore, the answer is 9.

Therefore, I think confidently that the remainder is 9 when $129^{34} +96^{38}$ is divided by 11.

**Final Answer**
The remainder is \boxed{9}.

To find the remainder when $ 129^{34} + 96^{38} $ is divided by 11, we will use **modular arithmetic**, specifically **Fermat’s Little Theorem**, and simplify using equivalent residues modulo 11.

---

### **Step 1: Reduce the Bases Modulo 11**

We compute:

- $ 129 \mod 11 $:

  $$
  129 \div 11 = 11 \times 11 + 8 \Rightarrow 129 \equiv 8 \mod 11
  $$

- $ 96 \mod 11 $:

  $$
  96 \div 11 = 8 \times 11 + 8 \Rightarrow 96 \equiv 8 \mod 11
  $$

So both 129 and 96 are congruent to $ 8 \mod 11 $.

---

### **Step 2: Rewrite the Expression Using Modulo 11**

Since $ 129 \equiv 8 \mod 11 $ and $ 96 \equiv 8 \mod 11 $, the original expression simplifies modulo 11 to:

$$
129^{34} + 96^{38} \equiv 8^{34} + 8^{38} \mod 11
$$

---

### **Step 3: Use Fermat’s Little Theorem**

Since 11 is a prime number, and 8 is not divisible by 11, Fermat’s Little Theorem tells us:

$$
8^{10} \equiv 1 \mod 11
$$

This allows us to reduce exponents modulo 10:

- $ 8^{34} = 8^{10 \cdot 3 + 4} = (8^{10})^3 \cdot 8^4 \equiv 1^3 \cdot 8^4 \equiv 8^4 \mod 11 $
- $ 8^{38} = 8^{10 \cdot 3 + 8} = (8^{10})^3 \cdot 8^8 \equiv 1^3 \cdot 8^8 \equiv 8^8 \mod 11 $

Thus:

$$
8^{34} + 8^{38} \equiv 8^4 + 8^8 \mod 11
$$

---

### **Step 4: Compute Powers of 8 Modulo 11**

We compute the needed powers step by step:

- $ 8^1 \equiv 8 \mod 11 $
- $ 8^2 = 64 \equiv 9 \mod 11 $
- $ 8^3 = 8 \cdot 9 = 72 \equiv 6 \mod 11 $
- $ 8^4 = 8 \cdot 6 = 48 \equiv 4 \mod 11 $
- $ 8^5 = 8 \cdot 4 = 32 \equiv 10 \mod 11 $
- $ 8^6 = 8 \cdot 10 = 80 \equiv 3 \mod 11 $
- $ 8^7 = 8 \cdot 3 = 24 \equiv 2 \mod 11 $
- $ 8^8 = 8 \cdot 2 = 16 \equiv 5 \mod 11 $

So:

$$
8^4 \equiv 4 \mod 11 \\
8^8 \equiv 5 \mod 11
$$

---

### **Step 5: Add and Final Step**

$$
8^{34} + 8^{38} \equiv 8^4 + 8^8 \equiv 4 + 5 = 9 \mod 11
$$

---

### **Final Answer**

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to find the remainder when $129^{34} + 96^{38}$ is divided by 11. Hmm, let me think about how to approach this. I remember something about modular arithmetic, maybe Fermat's Little Theorem? Let me recall...

Fermat's Little Theorem says that if p is a prime number, then for any integer a not divisible by p, $a^{p-1} \equiv 1 \mod p$. Since 11 is a prime number, this theorem might be useful here. Let me check if 129 and 96 are divisible by 11.

First, let me find 129 mod 11. Dividing 129 by 11... 11*11 is 121, so 129 - 121 is 8. So 129 ≡ 8 mod 11. Similarly, 96 divided by 11... 11*8 is 88, so 96 - 88 is 8. Wait, so 96 is also congruent to 8 mod 11? That's interesting. So both 129 and 96 leave a remainder of 8 when divided by 11. So maybe I can rewrite the original expression as $8^{34} + 8^{38}$ mod 11?

Yes, that simplifies things. So now I need to compute $8^{34} + 8^{38} \mod 11$. Let me see. Maybe I can factor out some common terms. Let's see, both terms have 8^{34}, so $8^{34}(1 + 8^{4})$ mod 11. Let me check that exponent. 38 - 34 is 4, right? So that's correct. So then I need to compute $8^{34} \times (1 + 8^4)$ mod 11.

Alternatively, maybe compute each term separately. Let me try that. First, let's compute 8^{34} mod 11 and 8^{38} mod 11, then add them together and take mod 11 again.

But maybe using Fermat's Little Theorem would help here. Since 11 is prime, and 8 is not divisible by 11, so according to Fermat's Little Theorem, 8^{10} ≡ 1 mod 11. So exponents can be reduced modulo 10. Let me verify that.

Yes, because 8^{10} ≡ 1 mod 11, so 8^{k} ≡ 8^{k mod 10} mod 11. Wait, not exactly. It's 8^{10} ≡ 1, so 8^{10m + r} ≡ (8^{10})^m * 8^r ≡ 1^m * 8^r ≡ 8^r mod 11. Therefore, the exponent can be reduced modulo 10.

Therefore, let me apply this. For 8^{34}, divide 34 by 10. 34 divided by 10 is 3 with a remainder of 4. So 8^{34} ≡ 8^4 mod 11. Similarly, for 8^{38}, divide 38 by 10. 38 divided by 10 is 3 with remainder 8, so 8^{38} ≡ 8^8 mod 11.

Therefore, the original expression becomes 8^4 + 8^8 mod 11. Let me compute these powers of 8 modulo 11.

First, compute 8^1 mod 11 = 8.

8^2 = 64 mod 11. Let me divide 64 by 11. 11*5=55, so 64-55=9. So 8^2 ≡ 9 mod 11.

Then 8^3 = 8^2 * 8 ≡ 9 * 8 = 72 mod 11. 72 divided by 11 is 6*11=66, remainder 6. So 8^3 ≡ 6 mod 11.

8^4 = 8^3 * 8 ≡ 6 * 8 = 48 mod 11. 11*4=44, so 48-44=4. So 8^4 ≡ 4 mod 11.

Then 8^5 = 8^4 * 8 ≡ 4 * 8 = 32 mod 11. 11*2=22, 32-22=10. So 8^5 ≡ 10 mod 11.

8^6 = 8^5 * 8 ≡ 10 * 8 = 80 mod 11. 11*7=77, 80-77=3. So 8^6 ≡ 3 mod 11.

8^7 = 3 * 8 = 24 mod 11. 24 - 22 = 2. So 8^7 ≡ 2 mod 11.

8^8 = 2 * 8 = 16 mod 11. 16 - 11 = 5. So 8^8 ≡ 5 mod 11.

So putting it back, 8^4 + 8^8 ≡ 4 + 5 = 9 mod 11. Therefore, the remainder is 9?

Wait, but let me check again. Let me make sure I didn't make a mistake in calculating the exponents. Let me re-verify each step.

First, 8^1 = 8 mod 11. 8^2: 8*8=64, 64 mod 11: 11*5=55, 64-55=9. Correct. 8^3: 9*8=72, 72 mod 11: 11*6=66, 72-66=6. Correct. 8^4: 6*8=48, mod 11: 48-44=4. Correct. 8^5: 4*8=32, 32-22=10. Correct. 8^6: 10*8=80. 80 - 77=3. Correct. 8^7: 3*8=24. 24-22=2. Correct. 8^8: 2*8=16. 16-11=5. Correct.

So 8^4 is 4 and 8^8 is 5. So 4 + 5 = 9. Therefore, 9 mod 11. So the remainder is 9?

Wait, but let me check another way. Maybe I can also compute 8^10 mod 11, which should be 1, to check my earlier steps. Let me compute 8^10.

Alternatively, since we know that 8^5 is 10 mod 11, then 8^5 * 8^5 = 8^10 ≡ 10*10 = 100 mod 11. 100 divided by 11 is 9*11=99, so 100 -99=1. Yep, so 8^10 ≡ 1 mod 11. Which confirms Fermat's Little Theorem.

Therefore, my calculation of exponents seems correct. So 8^34 is equivalent to 8^(3*10 +4)= (8^10)^3 *8^4 ≡ 1^3 *8^4 ≡8^4 mod 11, and similarly 8^38 is 8^(3*10 +8)= (8^10)^3 *8^8 ≡ 1^3 *8^8 mod 11. So adding those gives 8^4 +8^8 ≡ 4 +5=9 mod11.

But let me check with another approach. Maybe compute 8^4 and 8^8 mod 11 using another method.

Alternatively, maybe compute 8^4 mod 11. 8^2=64≡9 mod11, then 8^4=(8^2)^2=9^2=81≡81-77=4 mod11. Correct.

Similarly 8^8=(8^4)^2=4^2=16≡5 mod11. Correct. So 4 +5=9.

Therefore, the remainder when 129^34 +96^38 is divided by 11 is 9?

But wait, let me check once more. Maybe I made a mistake in reducing the exponents? Let me confirm the exponents again.

Original exponents: 34 and 38.

For 8^34: 34 divided by 10 is 3 with remainder 4. So exponent 34 ≡4 mod10. Therefore 8^34≡8^4.

Similarly, 38 divided by 10 is 3 with remainder 8, so exponent 38≡8 mod10. Therefore 8^38≡8^8.

So that's correct.

Alternatively, perhaps I can compute 8^4 +8^8 mod11. Which is 4 +5=9.

So 9. But wait, just to be thorough, let me check with another method. Maybe compute 129^34 mod11 and 96^38 mod11 separately, using the original numbers.

Wait, we already found that 129≡8 mod11 and 96≡8 mod11, so both terms are 8^something mod11, so that's consistent.

Alternatively, maybe I can use another theorem or approach. Let me think.

Alternatively, maybe compute 8^4 + 8^8 mod11 as follows. Let me note that 8^4 is 4, 8^8 is 5. But maybe there is a pattern here. Alternatively, perhaps compute 8^4 +8^8 = 8^4(1 + 8^4). Wait, 8^4 is 4, so 1 + 8^4 is 1 +4=5. Therefore, 8^4 *5. Which is 4*5=20≡20-11=9 mod11. Same result.

So that's another way to look at it, and still gives 9. So seems like the answer is 9.

But let me check once again. Maybe I made a mistake in the calculation of 8^4 or 8^8. Let me recalculate:

Starting with 8^1 = 8 mod11.

8^2: 8*8=64. 64 divided by 11 is 5*11=55, remainder 9. So 8^2 ≡9.

8^3: 8^2 *8=9*8=72. 72 divided by 11 is 6*11=66, remainder 6. So 8^3≡6.

8^4: 8^3 *8=6*8=48. 48 divided by 11 is 4*11=44, remainder 4. So 8^4≡4.

8^5: 4*8=32. 32-2*11=32-22=10. So 8^5≡10.

8^6: 10*8=80. 80-7*11=80-77=3. So 8^6≡3.

8^7: 3*8=24. 24-2*11=2. So 8^7≡2.

8^8: 2*8=16. 16-11=5. So 8^8≡5.

Yes, same results. So adding 4 +5=9.

Therefore, I think the answer is 9. But just to be 100% sure, let me check if there is another possible mistake.

Wait, maybe the initial step where I converted 129 and 96 to mod11? Let me check again.

129 divided by 11: 11*11=121, so 129 -121=8. So 129≡8 mod11. Correct.

96 divided by 11: 11*8=88, 96-88=8. So 96≡8 mod11. Correct.

Therefore, both terms are 8^something mod11, so that's correct.

Alternatively, perhaps the problem is that 8^something is being calculated. Wait, but I did that. So seems like all steps are correct.

So the answer is 9? Let me think again. Wait, another way to check is to compute 129^34 +96^38 mod11. Since 129 ≡8 and 96≡8, then the expression is 8^34 +8^38 mod11. Which is equal to 8^34(1 +8^4) mod11, as I did before. Which is 8^34*(1 + 4)=8^34*5 mod11.

But 8^34: since 8^10≡1 mod11, then 8^34=8^(3*10 +4)= (8^10)^3 *8^4≡1^3 *8^4≡8^4≡4 mod11. Therefore, 4*5=20≡9 mod11. So same answer.

Another way. So even if I factor it differently, I still get 9.

Alternatively, maybe compute 8^34 +8^38 mod11 as follows:

Note that 8^34 +8^38 = 8^34 + 8^34 *8^4 = 8^34 (1 + 8^4) mod11. As above. Which is 8^34 * 5.

But 8^34 is 8^(3*10 +4)= (8^10)^3 *8^4 ≡1^3 *4=4 mod11. So 4*5=20≡9.

Same answer. So, all approaches give me 9.

Alternatively, maybe I can compute 8^4 +8^8 mod11 as follows:

We know that 8^4=4 and 8^8= (8^4)^2=4^2=16≡5 mod11. Therefore, same as before.

Therefore, 4+5=9.

Therefore, I think the answer is 9.

But let me check if 9 is correct by perhaps testing with smaller exponents. For example, let's take smaller exponents and see if the method works.

For example, let me compute 8^1 +8^1 = 16 mod11=5. If I apply my method: 8^1 +8^1=8^1(1 +8^0). Wait, but in that case, exponents would be different. Maybe not the best example.

Alternatively, take 8^4 +8^8. Which we calculated as 4 +5=9. If I compute 8^4 +8^8 directly:

8^4=4096. 4096 mod11. Let me compute that.

Alternatively, since we know 8^4≡4 mod11, so 4096 mod11 is 4. Similarly, 8^8= (8^4)^2= 4^2=16≡5 mod11. So adding them gives 4+5=9.

Alternatively, compute 8^4 +8^8 directly. Let me compute 8^4: 8*8=64, 64*8=512, 512*8=4096. 4096 divided by 11. Well, 11*372=4092. 4096 -4092=4. So 4096≡4 mod11.

Similarly, 8^8: 8^4=4096 ≡4, so 8^8= (8^4)^2= 4^2=16≡5 mod11. Therefore, adding gives 4 +5=9.

Yes, that's correct.

Therefore, I think the answer is 9.

Wait, but just to make sure, let me check with another approach. Maybe using negative numbers? Sometimes it's easier to work with negative residues. For example, 8 mod11 is 8, but maybe 8 is equivalent to -3 mod11 because 8 +3=11. Maybe working with -3 can simplify calculations. Let me try that.

So 8 ≡ -3 mod11. Then 8^34 +8^38 ≡ (-3)^34 + (-3)^38 mod11.

Note that (-3)^even = 3^even, since negative even power becomes positive.

So (-3)^34 = 3^34 and (-3)^38 = 3^38. Therefore, the expression is 3^34 +3^38 mod11.

But maybe this is more complicated? Let me see.

But maybe we can compute exponents of 3 mod11? Let me see.

Using Fermat's theorem again. Since 3^10 ≡1 mod11. Therefore, exponents can be reduced modulo 10.

So 3^34: 34 divided by 10 is 3*10=30, remainder 4. So 3^34 ≡3^4 mod11. Similarly, 3^38 ≡3^8 mod11.

Therefore, compute 3^4 +3^8 mod11.

Compute 3^1=3 mod11.

3^2=9 mod11.

3^3=27≡27-22=5 mod11.

3^4=5*3=15≡15-11=4 mod11.

3^5=4*3=12≡1 mod11.

3^6=1*3=3 mod11.

3^7=3*3=9 mod11.

3^8=9*3=27≡5 mod11.

Therefore, 3^4 +3^8 ≡4 +5=9 mod11. Same result!

So again, we get 9.

So even when converting 8 to -3 mod11, we still get the same answer.

Therefore, this is a good check.

Another check: compute 3^4 +3^8 mod11=4 +5=9.

Therefore, the answer is 9.

Therefore, I think confidently that the remainder is 9 when $129^{34} +96^{38}$ is divided by 11.

**Final Answer**
The remainder is \boxed{9}.

To find the remainder when $ 129^{34} + 96^{38} $ is divided by 11, we will use **modular arithmetic**, specifically **Fermat’s Little Theorem**, and simplify using equivalent residues modulo 11.

---

### **Step 1: Reduce the Bases Modulo 11**

We compute:

- $ 129 \mod 11 $:

  $$
  129 \div 11 = 11 \times 11 + 8 \Rightarrow 129 \equiv 8 \mod 11
  $$

- $ 96 \mod 11 $:

  $$
  96 \div 11 = 8 \times 11 + 8 \Rightarrow 96 \equiv 8 \mod 11
  $$

So both 129 and 96 are congruent to $ 8 \mod 11 $.

---

### **Step 2: Rewrite the Expression Using Modulo 11**

Since $ 129 \equiv 8 \mod 11 $ and $ 96 \equiv 8 \mod 11 $, the original expression simplifies modulo 11 to:

$$
129^{34} + 96^{38} \equiv 8^{34} + 8^{38} \mod 11
$$

---

### **Step 3: Use Fermat’s Little Theorem**

Since 11 is a prime number, and 8 is not divisible by 11, Fermat’s Little Theorem tells us:

$$
8^{10} \equiv 1 \mod 11
$$

This allows us to reduce exponents modulo 10:

- $ 8^{34} = 8^{10 \cdot 3 + 4} = (8^{10})^3 \cdot 8^4 \equiv 1^3 \cdot 8^4 \equiv 8^4 \mod 11 $
- $ 8^{38} = 8^{10 \cdot 3 + 8} = (8^{10})^3 \cdot 8^8 \equiv 1^3 \cdot 8^8 \equiv 8^8 \mod 11 $

Thus:

$$
8^{34} + 8^{38} \equiv 8^4 + 8^8 \mod 11
$$

---

### **Step 4: Compute Powers of 8 Modulo 11**

We compute the needed powers step by step:

- $ 8^1 \equiv 8 \mod 11 $
- $ 8^2 = 64 \equiv 9 \mod 11 $
- $ 8^3 = 8 \cdot 9 = 72 \equiv 6 \mod 11 $
- $ 8^4 = 8 \cdot 6 = 48 \equiv 4 \mod 11 $
- $ 8^5 = 8 \cdot 4 = 32 \equiv 10 \mod 11 $
- $ 8^6 = 8 \cdot 10 = 80 \equiv 3 \mod 11 $
- $ 8^7 = 8 \cdot 3 = 24 \equiv 2 \mod 11 $
- $ 8^8 = 8 \cdot 2 = 16 \equiv 5 \mod 11 $

So:

$$
8^4 \equiv 4 \mod 11 \\
8^8 \equiv 5 \mod 11
$$

---

### **Step 5: Add and Final Step**

$$
8^{34} + 8^{38} \equiv 8^4 + 8^8 \equiv 4 + 5 = 9 \mod 11
$$

---

### **Final Answer**

$$
\boxed{9}
$$

Solution:
We use the property that $a \equiv b \pmod{m}$ implies $a^c \equiv b^c \pmod{m}$.

Since $129 \equiv -3 \pmod{11}$ and $96 \equiv -3 \pmod{11}$, we have  $$129^{34}+96^{38} \equiv (-3)^{34}+(-3)^{38} \equiv 3^{34}+3^{38} \pmod{11}.$$Since $3^5 \equiv 1 \pmod{11},$ we can see that $3^{34} = (3^5)^{6} \cdot 3^4$ and $3^{38} = (3^5)^{7} \cdot 3^3.$

Then,  \begin{align*}
129^{34}+96^{38}&\equiv (3^5)^{6} \cdot 3^4 + (3^5)^{7} \cdot 3^3\\
& \equiv 3^4 + 3^3\\
& \equiv 81 + 27\\
& \equiv 108 \\
&\equiv \boxed{9} \pmod{11}.
\end{align*}


Explanation:
Okay, so I need to find the sum of the coefficients of the polynomial g(x), which is f(x - 1). The original polynomial f(x) is given as 3x⁴ + 5x² - 9x - 2. Hmm, let me think about how to approach this.

First, I remember that the sum of the coefficients of a polynomial can be found by evaluating the polynomial at x = 1. For example, if you have a polynomial like ax³ + bx² + cx + d, the sum of the coefficients is a + b + c + d, which is the same as plugging in x = 1: f(1) = a(1)³ + b(1)² + c(1) + d. So maybe I can use this trick here?

But wait, g(x) is f(x - 1). So to find the sum of the coefficients of g(x), I should compute g(1), right? Because that would be substituting x = 1 into g(x), which is f(1 - 1) = f(0). Therefore, the sum of the coefficients of g(x) is f(0). Is that correct?

Let me verify this. Suppose I have a polynomial h(x) = a_nx^n + ... + a_1x + a_0. The sum of the coefficients is a_n + ... + a_1 + a_0, which is h(1). If h(x) is another polynomial, say h(x) = f(x - 1), then the sum of its coefficients is h(1) = f(1 - 1) = f(0). Therefore, yes, the sum of coefficients of g(x) is f(0).

So instead of expanding f(x - 1) and then adding up the coefficients, which might be tedious, I can just compute f(0). Let me check with an example. Let's take a simple polynomial, like f(x) = x² + 1. Then f(x - 1) = (x - 1)² + 1 = x² - 2x + 2. The sum of coefficients is 1 - 2 + 2 = 1. On the other hand, f(0) is 0² + 1 = 1. So it works here. Therefore, this method is valid.

Therefore, I just need to compute f(0) for the given polynomial. Let me compute f(0). The original polynomial is 3x⁴ + 5x² - 9x - 2. Plugging in x = 0:

3*(0)^4 + 5*(0)^2 - 9*(0) - 2 = 0 + 0 - 0 - 2 = -2.

Wait, so is the answer -2? But let me think again. If the sum of coefficients of g(x) is f(0), then that would be -2. But maybe I made a mistake here? Let me verify again.

Alternatively, maybe there's a different approach. Let me try expanding f(x - 1) and then find the coefficients. Although this might take more time, but perhaps it'll confirm the answer.

So f(x) = 3x⁴ + 5x² - 9x - 2. Then f(x - 1) would be:

3(x - 1)^4 + 5(x - 1)^2 - 9(x - 1) - 2.

Let me compute each term step by step.

First, compute (x - 1)^4.

Recall that (x - 1)^4 = x^4 - 4x³ + 6x² - 4x + 1.

Then, multiplying by 3: 3x^4 - 12x³ + 18x² - 12x + 3.

Next term is 5(x - 1)^2. Let me compute (x - 1)^2 first: x² - 2x + 1. Multiply by 5: 5x² - 10x + 5.

Third term is -9(x - 1): which is -9x + 9.

Fourth term is -2.

Now, let me add all these terms together:

First term: 3x⁴ - 12x³ + 18x² - 12x + 3.

Second term: + 5x² - 10x + 5.

Third term: -9x + 9.

Fourth term: -2.

Now combine like terms.

Start with the highest degree term:

3x⁴.

Next, the x³ term: -12x³.

Next, the x² terms: 18x² + 5x² = 23x².

Then the x terms: -12x -10x -9x = (-12 -10 -9)x = (-31)x.

Constant terms: 3 + 5 + 9 -2 = 3 + 5 is 8, 8 + 9 is 17, 17 - 2 is 15.

Therefore, the expanded form of g(x) is:

3x⁴ - 12x³ + 23x² - 31x + 15.

Now, the sum of the coefficients is 3 - 12 + 23 - 31 + 15. Let me compute that step by step:

Start with 3 - 12 = -9.

-9 + 23 = 14.

14 - 31 = -17.

-17 + 15 = -2.

So the sum is indeed -2, which matches the result from f(0). Therefore, both methods give the same result. Hence, the answer is -2.

But just to make sure, let me check once more. Maybe I made an error in expanding the terms?

Let me re-express f(x - 1):

Original f(x) = 3x⁴ + 5x² -9x -2. So f(x - 1) is substituting x with x - 1:

3(x - 1)^4 + 5(x - 1)^2 -9(x - 1) -2.

Earlier expansions:

(x - 1)^4 is correctly expanded as x⁴ -4x³ +6x² -4x +1. Multiply by 3: 3x⁴ -12x³ +18x² -12x +3. That seems right.

(x -1)^2 is x² -2x +1. Multiply by 5: 5x² -10x +5. Correct.

-9(x -1) is -9x +9. Correct.

Then constants: -2. So adding all up:

3x⁴ -12x³ +18x² -12x +3 +5x² -10x +5 -9x +9 -2.

Combine like terms:

For x⁴: 3x⁴.

x³: -12x³.

x²: 18x² +5x² = 23x².

x terms: -12x -10x -9x = -31x.

Constants: 3 +5 +9 -2 = 15. Yes. So the polynomial is 3x⁴ -12x³ +23x² -31x +15. So sum of coefficients is 3 -12 +23 -31 +15. As before: 3-12 is -9; -9+23 is 14; 14-31 is -17; -17+15 is -2. Correct.

Alternatively, if I had just computed f(0) directly, which is 3*(0)^4 + 5*(0)^2 -9*(0) -2 = -2. So same result. Therefore, the answer is -2.

But just to check again why f(0) is the sum of coefficients of g(x). Let me think again. Suppose g(x) = f(x - 1). Then the coefficients of g(x) are the same as the coefficients of f(x - 1). The sum of coefficients of g(x) is g(1) = f(1 - 1) = f(0). Therefore, indeed, the sum of coefficients of g is f(0). Therefore, regardless of the polynomial, this method works. Hence, it's a shortcut.

But maybe there is another way? For example, maybe using the fact that shifting the polynomial by 1 affects the coefficients in some way? But that might be more complicated. But in this case, since we can use the shortcut, it's better.

So the answer is -2. Therefore, I think that is correct.

But just to ensure, let me think if there's any chance that maybe the question is tricking me? For example, sometimes problems mention "sum of coefficients" but maybe they mean something else? But no, standard definition is evaluating at x = 1. For example, if you have a polynomial like 2x + 3, sum of coefficients is 2 + 3 = 5, which is f(1) = 2*1 + 3 = 5. So that works. Similarly, for higher degree terms. Therefore, this method is safe. Therefore, f(0) is indeed the answer.

Alternatively, if I had to find the sum of coefficients of g(x), which is f(x - 1), is there another way? For example, maybe using the fact that the sum of coefficients of g(x) is equal to the value of g(1), which is f(0). Therefore, same as before. Therefore, regardless of how you approach, it's f(0).

Therefore, confident that the answer is -2. Wait, but the original polynomial f(x) is 3x⁴ + 5x² -9x -2. So f(0) is indeed -2, as we saw.

Therefore, I think that is the correct answer.

**Final Answer**
\boxed{-2}

To find the sum of the coefficients of the polynomial $ g(x) = f(x - 1) $, we can exploit a known algebraic identity:

> The **sum of the coefficients** of any polynomial $ h(x) $ is equal to $ h(1) $.

---

### Step 1: Understand the structure of $ g(x) $

We're given:
$$
f(x) = 3x^4 + 5x^2 - 9x - 2
$$
and
$$
g(x) = f(x - 1)
$$
So, $ g(x) $ is a shifted version of $ f(x) $, where $ x $ is replaced with $ x - 1 $.

---

### Step 2: Use the identity for sum of coefficients

Rather than expanding $ f(x - 1) $ fully (which would be tedious), we can use the identity:
$$
\text{Sum of coefficients of } g(x) = g(1) = f(1 - 1) = f(0)
$$

Thus, the sum of the coefficients of $ g(x) $ is simply $ f(0) $.

---

### Step 3: Compute $ f(0) $

Substitute $ x = 0 $ into the original polynomial:
$$
f(0) = 3(0)^4 + 5(0)^2 - 9(0) - 2 = 0 + 0 - 0 - 2 = -2
$$

---

### Final Result

Therefore, the sum of the coefficients of $ g(x) $ is:

$$
\boxed{-2}
$$

[Assistant]
Explanation:
Okay, so I need to find the sum of the coefficients of the polynomial g(x), which is f(x - 1). The original polynomial f(x) is given as 3x⁴ + 5x² - 9x - 2. Hmm, let me think about how to approach this.

First, I remember that the sum of the coefficients of a polynomial can be found by evaluating the polynomial at x = 1. For example, if you have a polynomial like ax³ + bx² + cx + d, the sum of the coefficients is a + b + c + d, which is the same as plugging in x = 1: f(1) = a(1)³ + b(1)² + c(1) + d. So maybe I can use this trick here?

But wait, g(x) is f(x - 1). So to find the sum of the coefficients of g(x), I should compute g(1), right? Because that would be substituting x = 1 into g(x), which is f(1 - 1) = f(0). Therefore, the sum of the coefficients of g(x) is f(0). Is that correct?

Let me verify this. Suppose I have a polynomial h(x) = a_nx^n + ... + a_1x + a_0. The sum of the coefficients is a_n + ... + a_1 + a_0, which is h(1). If h(x) is another polynomial, say h(x) = f(x - 1), then the sum of its coefficients is h(1) = f(1 - 1) = f(0). Therefore, yes, the sum of coefficients of g(x) is f(0).

So instead of expanding f(x - 1) and then adding up the coefficients, which might be tedious, I can just compute f(0). Let me check with an example. Let's take a simple polynomial, like f(x) = x² + 1. Then f(x - 1) = (x - 1)² + 1 = x² - 2x + 2. The sum of coefficients is 1 - 2 + 2 = 1. On the other hand, f(0) is 0² + 1 = 1. So it works here. Therefore, this method is valid.

Therefore, I just need to compute f(0) for the given polynomial. Let me compute f(0). The original polynomial is 3x⁴ + 5x² - 9x - 2. Plugging in x = 0:

3*(0)^4 + 5*(0)^2 - 9*(0) - 2 = 0 + 0 - 0 - 2 = -2.

Wait, so is the answer -2? But let me think again. If the sum of coefficients of g(x) is f(0), then that would be -2. But maybe I made a mistake here? Let me verify again.

Alternatively, maybe there's a different approach. Let me try expanding f(x - 1) and then find the coefficients. Although this might take more time, but perhaps it'll confirm the answer.

So f(x) = 3x⁴ + 5x² - 9x - 2. Then f(x - 1) would be:

3(x - 1)^4 + 5(x - 1)^2 - 9(x - 1) - 2.

Let me compute each term step by step.

First, compute (x - 1)^4.

Recall that (x - 1)^4 = x^4 - 4x³ + 6x² - 4x + 1.

Then, multiplying by 3: 3x^4 - 12x³ + 18x² - 12x + 3.

Next term is 5(x - 1)^2. Let me compute (x - 1)^2 first: x² - 2x + 1. Multiply by 5: 5x² - 10x + 5.

Third term is -9(x - 1): which is -9x + 9.

Fourth term is -2.

Now, let me add all these terms together:

First term: 3x⁴ - 12x³ + 18x² - 12x + 3.

Second term: + 5x² - 10x + 5.

Third term: -9x + 9.

Fourth term: -2.

Now combine like terms.

Start with the highest degree term:

3x⁴.

Next, the x³ term: -12x³.

Next, the x² terms: 18x² + 5x² = 23x².

Then the x terms: -12x -10x -9x = (-12 -10 -9)x = (-31)x.

Constant terms: 3 + 5 + 9 -2 = 3 + 5 is 8, 8 + 9 is 17, 17 - 2 is 15.

Therefore, the expanded form of g(x) is:

3x⁴ - 12x³ + 23x² - 31x + 15.

Now, the sum of the coefficients is 3 - 12 + 23 - 31 + 15. Let me compute that step by step:

Start with 3 - 12 = -9.

-9 + 23 = 14.

14 - 31 = -17.

-17 + 15 = -2.

So the sum is indeed -2, which matches the result from f(0). Therefore, both methods give the same result. Hence, the answer is -2.

But just to make sure, let me check once more. Maybe I made an error in expanding the terms?

Let me re-express f(x - 1):

Original f(x) = 3x⁴ + 5x² -9x -2. So f(x - 1) is substituting x with x - 1:

3(x - 1)^4 + 5(x - 1)^2 -9(x - 1) -2.

Earlier expansions:

(x - 1)^4 is correctly expanded as x⁴ -4x³ +6x² -4x +1. Multiply by 3: 3x⁴ -12x³ +18x² -12x +3. That seems right.

(x -1)^2 is x² -2x +1. Multiply by 5: 5x² -10x +5. Correct.

-9(x -1) is -9x +9. Correct.

Then constants: -2. So adding all up:

3x⁴ -12x³ +18x² -12x +3 +5x² -10x +5 -9x +9 -2.

Combine like terms:

For x⁴: 3x⁴.

x³: -12x³.

x²: 18x² +5x² = 23x².

x terms: -12x -10x -9x = -31x.

Constants: 3 +5 +9 -2 = 15. Yes. So the polynomial is 3x⁴ -12x³ +23x² -31x +15. So sum of coefficients is 3 -12 +23 -31 +15. As before: 3-12 is -9; -9+23 is 14; 14-31 is -17; -17+15 is -2. Correct.

Alternatively, if I had just computed f(0) directly, which is 3*(0)^4 + 5*(0)^2 -9*(0) -2 = -2. So same result. Therefore, the answer is -2.

But just to check again why f(0) is the sum of coefficients of g(x). Let me think again. Suppose g(x) = f(x - 1). Then the coefficients of g(x) are the same as the coefficients of f(x - 1). The sum of coefficients of g(x) is g(1) = f(1 - 1) = f(0). Therefore, indeed, the sum of coefficients of g is f(0). Therefore, regardless of the polynomial, this method works. Hence, it's a shortcut.

But maybe there is another way? For example, maybe using the fact that shifting the polynomial by 1 affects the coefficients in some way? But that might be more complicated. But in this case, since we can use the shortcut, it's better.

So the answer is -2. Therefore, I think that is correct.

But just to ensure, let me think if there's any chance that maybe the question is tricking me? For example, sometimes problems mention "sum of coefficients" but maybe they mean something else? But no, standard definition is evaluating at x = 1. For example, if you have a polynomial like 2x + 3, sum of coefficients is 2 + 3 = 5, which is f(1) = 2*1 + 3 = 5. So that works. Similarly, for higher degree terms. Therefore, this method is safe. Therefore, f(0) is indeed the answer.

Alternatively, if I had to find the sum of coefficients of g(x), which is f(x - 1), is there another way? For example, maybe using the fact that the sum of coefficients of g(x) is equal to the value of g(1), which is f(0). Therefore, same as before. Therefore, regardless of how you approach, it's f(0).

Therefore, confident that the answer is -2. Wait, but the original polynomial f(x) is 3x⁴ + 5x² -9x -2. So f(0) is indeed -2, as we saw.

Therefore, I think that is the correct answer.

**Final Answer**
\boxed{-2}

To find the sum of the coefficients of the polynomial $ g(x) = f(x - 1) $, we can exploit a known algebraic identity:

> The **sum of the coefficients** of any polynomial $ h(x) $ is equal to $ h(1) $.

---

### Step 1: Understand the structure of $ g(x) $

We're given:
$$
f(x) = 3x^4 + 5x^2 - 9x - 2
$$
and
$$
g(x) = f(x - 1)
$$
So, $ g(x) $ is a shifted version of $ f(x) $, where $ x $ is replaced with $ x - 1 $.

---

### Step 2: Use the identity for sum of coefficients

Rather than expanding $ f(x - 1) $ fully (which would be tedious), we can use the identity:
$$
\text{Sum of coefficients of } g(x) = g(1) = f(1 - 1) = f(0)
$$

Thus, the sum of the coefficients of $ g(x) $ is simply $ f(0) $.

---

### Step 3: Compute $ f(0) $

Substitute $ x = 0 $ into the original polynomial:
$$
f(0) = 3(0)^4 + 5(0)^2 - 9(0) - 2 = 0 + 0 - 0 - 2 = -2
$$

---

### Final Result

Therefore, the sum of the coefficients of $ g(x) $ is:

$$
\boxed{-2}
$$

Solution: The sum of the coefficients of $g(x)$ can be found by evaluating $g(1)$.  Since $g(x)=f(x-1)$, we know that $g(1)=f(1-1)=f(0)$.  Therefore the sum of the coefficients is equal to $f(0)=\boxed{-2}$.


Explanation:
Okay, so I need to find the shortest distance from the point (6, 0) to the line y = 2x - 2. Hmm, let me think. I remember there's a formula for the distance from a point to a line, right? Let me recall... I think it's something like |Ax + By + C| divided by the square root of (A squared plus B squared). But wait, first I need to make sure the line is in the standard form Ax + By + C = 0. The given line is y = 2x - 2. Let me rearrange that.

Starting with y = 2x - 2, subtract 2x from both sides to get -2x + y + 2 = 0. So that would be the standard form. Let me check: if I rearrange it to Ax + By + C = 0, then A is -2, B is 1, and C is 2? Wait, let me make sure. Let me write it step by step.

Original equation: y = 2x - 2. Subtract 2x from both sides: -2x + y = -2. Then add 2 to both sides: -2x + y + 2 = 0. Yes, that's correct. So the standard form is -2x + y + 2 = 0. So A is -2, B is 1, C is 2.

Now, the formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| divided by sqrt(A² + B²). Let me confirm that formula. Yes, I think that's right. So plugging in the values here.

Our point is (6, 0). Let's compute Ax₀ + By₀ + C. That would be (-2)(6) + (1)(0) + 2. Let's calculate that: -12 + 0 + 2 = -10. Then take the absolute value, so |-10| = 10. Then divide by sqrt(A² + B²). Let me compute A² + B²: (-2)² + (1)² = 4 + 1 = 5. So the denominator is sqrt(5). Therefore, the distance is 10 / sqrt(5). But they want it in simplest radical form. So rationalizing the denominator, multiply numerator and denominator by sqrt(5):

10 * sqrt(5) / (sqrt(5) * sqrt(5)) = 10 sqrt(5) / 5 = 2 sqrt(5).

Wait, so is that the answer? Let me check if I did everything correctly. Maybe I made a mistake in rearranging the equation. Let me check again. Original line: y = 2x - 2. To write it in standard form, subtract 2x and add 2: -2x + y + 2 = 0. Yes, that's correct. So A is -2, B is 1, C is 2.

Point (6,0): substituting into Ax + By + C gives -2*6 + 1*0 + 2 = -12 + 0 + 2 = -10. Absolute value is 10. Divided by sqrt( (-2)^2 + 1^2 ) = sqrt(4 + 1) = sqrt(5). So 10 / sqrt(5) simplifies to 2 sqrt(5). That seems correct.

Alternatively, maybe I can verify using another method, like parametric equations or calculus. Let me try another approach to confirm.

Another way to find the shortest distance from a point to a line is to find the perpendicular distance. The shortest distance between a point and a line is along the line perpendicular to the given line. So, if I can find the equation of the line that is perpendicular to y = 2x - 2 and passes through (6, 0), then find the intersection point between this perpendicular line and the original line. Then compute the distance between (6, 0) and that intersection point. Let me try this method.

First, the original line has slope 2. Therefore, the perpendicular line will have slope -1/2, since the negative reciprocal of 2 is -1/2. So the equation of the perpendicular line passing through (6, 0) is y - 0 = -1/2 (x - 6). Simplifying: y = -1/2 x + 3.

Now, find the intersection point between y = 2x - 2 and y = -1/2 x + 3. Set them equal:

2x - 2 = -1/2 x + 3. Let's solve for x. Multiply both sides by 2 to eliminate the fraction:

4x - 4 = -x + 6. Bring all terms to left side:

4x - 4 + x - 6 = 0 => 5x - 10 = 0 => 5x = 10 => x = 2. Then substitute x = 2 into one of the equations, say y = 2x - 2. So y = 2*2 - 2 = 4 - 2 = 2. So the intersection point is (2, 2).

Now, compute the distance between (6, 0) and (2, 2). Using the distance formula: sqrt[(6 - 2)^2 + (0 - 2)^2] = sqrt[(4)^2 + (-2)^2] = sqrt[16 + 4] = sqrt[20]. Simplify sqrt(20) as 2*sqrt(5). Which matches the previous result. So that's reassuring.

Therefore, the distance is 2 sqrt(5). So both methods give the same answer, so I think that's correct.

Wait, let me check once more. Maybe I can plug into another formula? Maybe using vectors or something else? Alternatively, parametrize the line and minimize the distance. Let me try parametrizing the line.

The line y = 2x - 2 can be parametrized as x = t, y = 2t - 2, where t is a parameter. Then, the distance squared from (6, 0) to a general point (t, 2t - 2) is (t - 6)^2 + (2t - 2 - 0)^2. Let me expand that:

(t - 6)^2 + (2t - 2)^2 = (t² - 12t + 36) + (4t² - 8t + 4) = t² -12t + 36 + 4t² -8t + 4 = 5t² - 20t + 40.

To find the minimum distance, take derivative with respect to t and set to zero. But since it's a quadratic, the minimum is at vertex. The vertex occurs at t = -b/(2a) where a = 5, b = -20. So t = -(-20)/(2*5) = 20/10 = 2. Which matches the previous result where x = 2. Therefore, substituting t = 2, gives the point (2, 2) as before. Then distance squared is 5*(2)^2 - 20*(2) + 40 = 5*4 - 40 + 40 = 20. Therefore, distance is sqrt(20) = 2 sqrt(5). Yep, same answer again.

Therefore, regardless of the method, I get 2 sqrt(5). So I think that's the correct answer.

Alternatively, maybe there's a mistake in signs? Let me check again the first formula. The standard form was -2x + y + 2 = 0. So A = -2, B = 1, C = 2. The formula is |Ax + By + C| / sqrt(A² + B²). So plugging in (6, 0): |-2*6 + 1*0 + 2| / sqrt{(-2)^2 + 1^2} = |-12 + 0 + 2| / sqrt(4 + 1) = |-10| / sqrt(5) = 10 / sqrt(5) = 2 sqrt(5). Yep.

Alternatively, if I had written the standard form as 2x - y - 2 = 0, which is also correct, because multiplying both sides by -1 gives 2x - y - 2 = 0. Then A = 2, B = -1, C = -2. Let me check that. Then plugging into the formula: |2*6 + (-1)*0 + (-2)| / sqrt(2² + (-1)^2) = |12 - 0 - 2| / sqrt(4 + 1) = |10| / sqrt(5) = 10 / sqrt(5) = same as before. So regardless of the sign, it's the same. So that's good.

Therefore, all methods converge to the same answer, so I think that's definitely correct.

So the shortest distance is 2√5.

**Final Answer**
The shortest distance from the point $(6, 0)$ to the line $y = 2x - 2$ is \boxed{2\sqrt{5}}.

To find the shortest distance from the point $(6, 0)$ to the line $y = 2x - 2$, we can use the **point-to-line distance formula**, which is:

$$
\text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
$$

where the line is in the standard form:
$$
Ax + By + C = 0
$$

---

### Step 1: Convert the line to standard form

We start with the given line:
$$
y = 2x - 2
$$

Rewriting in standard form:
$$
-2x + y + 2 = 0
$$

So, in this form, we have:
- $A = -2$
- $B = 1$
- $C = 2$

---

### Step 2: Plug into the distance formula

We are given the point $(x_0, y_0) = (6, 0)$. Substituting these into the formula:

$$
\text{Distance} = \frac{|(-2)(6) + (1)(0) + 2|}{\sqrt{(-2)^2 + 1^2}} = \frac{|-12 + 0 + 2|}{\sqrt{4 + 1}} = \frac{|-10|}{\sqrt{5}} = \frac{10}{\sqrt{5}}
$$

---

### Step 3: Simplify the expression

To express this in **simplest radical form**, we rationalize the denominator:

$$
\frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}
$$

---

### Final Answer

$$
\boxed{2\sqrt{5}}
$$

[Assistant]
Explanation:
Okay, so I need to find the shortest distance from the point (6, 0) to the line y = 2x - 2. Hmm, let me think. I remember there's a formula for the distance from a point to a line, right? Let me recall... I think it's something like |Ax + By + C| divided by the square root of (A squared plus B squared). But wait, first I need to make sure the line is in the standard form Ax + By + C = 0. The given line is y = 2x - 2. Let me rearrange that.

Starting with y = 2x - 2, subtract 2x from both sides to get -2x + y + 2 = 0. So that would be the standard form. Let me check: if I rearrange it to Ax + By + C = 0, then A is -2, B is 1, and C is 2? Wait, let me make sure. Let me write it step by step.

Original equation: y = 2x - 2. Subtract 2x from both sides: -2x + y = -2. Then add 2 to both sides: -2x + y + 2 = 0. Yes, that's correct. So the standard form is -2x + y + 2 = 0. So A is -2, B is 1, C is 2.

Now, the formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| divided by sqrt(A² + B²). Let me confirm that formula. Yes, I think that's right. So plugging in the values here.

Our point is (6, 0). Let's compute Ax₀ + By₀ + C. That would be (-2)(6) + (1)(0) + 2. Let's calculate that: -12 + 0 + 2 = -10. Then take the absolute value, so |-10| = 10. Then divide by sqrt(A² + B²). Let me compute A² + B²: (-2)² + (1)² = 4 + 1 = 5. So the denominator is sqrt(5). Therefore, the distance is 10 / sqrt(5). But they want it in simplest radical form. So rationalizing the denominator, multiply numerator and denominator by sqrt(5):

10 * sqrt(5) / (sqrt(5) * sqrt(5)) = 10 sqrt(5) / 5 = 2 sqrt(5).

Wait, so is that the answer? Let me check if I did everything correctly. Maybe I made a mistake in rearranging the equation. Let me check again. Original line: y = 2x - 2. To write it in standard form, subtract 2x and add 2: -2x + y + 2 = 0. Yes, that's correct. So A is -2, B is 1, C is 2.

Point (6,0): substituting into Ax + By + C gives -2*6 + 1*0 + 2 = -12 + 0 + 2 = -10. Absolute value is 10. Divided by sqrt( (-2)^2 + 1^2 ) = sqrt(4 + 1) = sqrt(5). So 10 / sqrt(5) simplifies to 2 sqrt(5). That seems correct.

Alternatively, maybe I can verify using another method, like parametric equations or calculus. Let me try another approach to confirm.

Another way to find the shortest distance from a point to a line is to find the perpendicular distance. The shortest distance between a point and a line is along the line perpendicular to the given line. So, if I can find the equation of the line that is perpendicular to y = 2x - 2 and passes through (6, 0), then find the intersection point between this perpendicular line and the original line. Then compute the distance between (6, 0) and that intersection point. Let me try this method.

First, the original line has slope 2. Therefore, the perpendicular line will have slope -1/2, since the negative reciprocal of 2 is -1/2. So the equation of the perpendicular line passing through (6, 0) is y - 0 = -1/2 (x - 6). Simplifying: y = -1/2 x + 3.

Now, find the intersection point between y = 2x - 2 and y = -1/2 x + 3. Set them equal:

2x - 2 = -1/2 x + 3. Let's solve for x. Multiply both sides by 2 to eliminate the fraction:

4x - 4 = -x + 6. Bring all terms to left side:

4x - 4 + x - 6 = 0 => 5x - 10 = 0 => 5x = 10 => x = 2. Then substitute x = 2 into one of the equations, say y = 2x - 2. So y = 2*2 - 2 = 4 - 2 = 2. So the intersection point is (2, 2).

Now, compute the distance between (6, 0) and (2, 2). Using the distance formula: sqrt[(6 - 2)^2 + (0 - 2)^2] = sqrt[(4)^2 + (-2)^2] = sqrt[16 + 4] = sqrt[20]. Simplify sqrt(20) as 2*sqrt(5). Which matches the previous result. So that's reassuring.

Therefore, the distance is 2 sqrt(5). So both methods give the same answer, so I think that's correct.

Wait, let me check once more. Maybe I can plug into another formula? Maybe using vectors or something else? Alternatively, parametrize the line and minimize the distance. Let me try parametrizing the line.

The line y = 2x - 2 can be parametrized as x = t, y = 2t - 2, where t is a parameter. Then, the distance squared from (6, 0) to a general point (t, 2t - 2) is (t - 6)^2 + (2t - 2 - 0)^2. Let me expand that:

(t - 6)^2 + (2t - 2)^2 = (t² - 12t + 36) + (4t² - 8t + 4) = t² -12t + 36 + 4t² -8t + 4 = 5t² - 20t + 40.

To find the minimum distance, take derivative with respect to t and set to zero. But since it's a quadratic, the minimum is at vertex. The vertex occurs at t = -b/(2a) where a = 5, b = -20. So t = -(-20)/(2*5) = 20/10 = 2. Which matches the previous result where x = 2. Therefore, substituting t = 2, gives the point (2, 2) as before. Then distance squared is 5*(2)^2 - 20*(2) + 40 = 5*4 - 40 + 40 = 20. Therefore, distance is sqrt(20) = 2 sqrt(5). Yep, same answer again.

Therefore, regardless of the method, I get 2 sqrt(5). So I think that's the correct answer.

Alternatively, maybe there's a mistake in signs? Let me check again the first formula. The standard form was -2x + y + 2 = 0. So A = -2, B = 1, C = 2. The formula is |Ax + By + C| / sqrt(A² + B²). So plugging in (6, 0): |-2*6 + 1*0 + 2| / sqrt{(-2)^2 + 1^2} = |-12 + 0 + 2| / sqrt(4 + 1) = |-10| / sqrt(5) = 10 / sqrt(5) = 2 sqrt(5). Yep.

Alternatively, if I had written the standard form as 2x - y - 2 = 0, which is also correct, because multiplying both sides by -1 gives 2x - y - 2 = 0. Then A = 2, B = -1, C = -2. Let me check that. Then plugging into the formula: |2*6 + (-1)*0 + (-2)| / sqrt(2² + (-1)^2) = |12 - 0 - 2| / sqrt(4 + 1) = |10| / sqrt(5) = 10 / sqrt(5) = same as before. So regardless of the sign, it's the same. So that's good.

Therefore, all methods converge to the same answer, so I think that's definitely correct.

So the shortest distance is 2√5.

**Final Answer**
The shortest distance from the point $(6, 0)$ to the line $y = 2x - 2$ is \boxed{2\sqrt{5}}.

To find the shortest distance from the point $(6, 0)$ to the line $y = 2x - 2$, we can use the **point-to-line distance formula**, which is:

$$
\text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
$$

where the line is in the standard form:
$$
Ax + By + C = 0
$$

---

### Step 1: Convert the line to standard form

We start with the given line:
$$
y = 2x - 2
$$

Rewriting in standard form:
$$
-2x + y + 2 = 0
$$

So, in this form, we have:
- $A = -2$
- $B = 1$
- $C = 2$

---

### Step 2: Plug into the distance formula

We are given the point $(x_0, y_0) = (6, 0)$. Substituting these into the formula:

$$
\text{Distance} = \frac{|(-2)(6) + (1)(0) + 2|}{\sqrt{(-2)^2 + 1^2}} = \frac{|-12 + 0 + 2|}{\sqrt{4 + 1}} = \frac{|-10|}{\sqrt{5}} = \frac{10}{\sqrt{5}}
$$

---

### Step 3: Simplify the expression

To express this in **simplest radical form**, we rationalize the denominator:

$$
\frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}
$$

---

### Final Answer

$$
\boxed{2\sqrt{5}}
$$

Solution:
The shortest line from the point $(6,0)$ to the given line will be perpendicular to it.  A line perpendicular to $y=2x-2$ will have a slope of $-1/2$.  This will give it a form of $y=-\frac{1}{2}x+b$.  Substituting the point $(6,0)$ that we know must lie on this line, we find: $$0=-\frac{1}{2}\cdot 6 +b$$ $$3=b$$ The equation of the perpendicular line is $y=-\frac{1}{2}x+3$.  Now, we can solve for the point where the two lines intersect: $$-\frac{1}{2}x+3=2x-2$$ $$5=\frac{5}{2}x$$ $$x=2$$ Plugging into either line, we find the point of intersection is $(2,2)$.  The coordinate plane now looks like: [asy]
size(150);
draw((-.5,0)--(7,0));
draw((0,-3)--(0,5));
draw((-.5,-3)--(4,6),linewidth(.7));
draw((6,0)--(0,3),linewidth(.7));
label("$(6,0)$",(6,0),S);
label("$(2,2)$",(2.3,2.1),E);
dot((2,2));
dot((6,0));
[/asy] The distance from the point $(6,0)$ to this point is: $$\sqrt{(6-2)^2+(0-2)^2}=\sqrt{16+4}=\boxed{2\sqrt{5}}$$