Explanation:
Okay, so I need to figure out where the reflection takes the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$. The problem says 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. If I can find that matrix, then I can apply it to any vector, including $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$, to find its image under the reflection.

First, let me remember that a reflection matrix in 2D can be represented in different ways depending on the line of reflection. The most common ones are reflections over the x-axis, y-axis, or the line y = x. But since the problem doesn't specify the line, I need to figure out which line is being reflected over based on the given information.

So, given that reflecting $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ gives $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, I can use this information to determine the reflection line. Let me think. If I can find the line of reflection, then I can write the reflection matrix accordingly.

Alternatively, since reflections are linear transformations, maybe I can set up equations to find the reflection matrix. Let me try that approach.

Let me denote the reflection matrix as $ R = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $. Since it's a reflection, it should satisfy certain properties. For example, it should be its own inverse, meaning $ R^2 = I $, and its determinant should be -1. Also, the trace of the matrix should be 0 because for a reflection matrix, trace is 0. Wait, is that true?

Wait, let me check. For a reflection over a line making an angle θ with the x-axis, the reflection matrix is:

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

So, the trace of this matrix is $ \cos 2\theta - \cos 2\theta = 0 $. So yes, the trace is 0. Also, determinant is $ -\cos^2 2\theta - \sin^2 2\theta $? Wait, no. Let me compute determinant:

Determinant of R is $ (a)(d) - (b)(c) $. For the matrix above, that would be $ \cos 2\theta \times (-\cos 2\theta) - \sin 2\theta \times \sin 2\theta = -\cos^2 2\theta - \sin^2 2\theta = -(\cos^2 2\theta + \sin^2 2\theta) = -1 $. So determinant is -1. So, for any reflection matrix, determinant is -1 and trace is 0. So, if I can find such a matrix R that satisfies R * v = w, where v is $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ and w is $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, then I can find the matrix.

So, let me write the equation:

$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Multiplying the matrix with the vector:

First component: 5a + 0*b = 5a = 4 ⇒ a = 4/5

Second component: 5c + 0*d = 5c = 3 ⇒ c = 3/5

So, we have a = 4/5 and c = 3/5. Now, since the matrix is a reflection matrix, we know that trace is 0, so a + d = 0 ⇒ d = -a = -4/5.

Also, determinant is -1, so ad - bc = -1. Let me check that.

We have a = 4/5, d = -4/5, c = 3/5. So, ad - bc = (4/5)(-4/5) - b*(3/5) = -16/25 - (3b)/5 = -1.

So, solving for b:

-16/25 - (3b)/5 = -1 ⇒ - (3b)/5 = -1 + 16/25 = (-25/25 + 16/25) = -9/25 ⇒

Multiply both sides by -1: (3b)/5 = 9/25 ⇒ 3b = 9/5 ⇒ b = 3/5.

So, b = 3/5. Therefore, the reflection matrix is:

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

Let me check if this matrix satisfies the determinant condition. ad - bc = (4/5)(-4/5) - (3/5)(3/5) = -16/25 - 9/25 = -25/25 = -1. Yes, that works. And trace is 4/5 + (-4/5) = 0. Perfect.

So, now that I have the reflection matrix, I can apply it to the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to find its image.

Let me compute that:

$$
R \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \begin{pmatrix} (4/5)(-2) + (3/5)(3) \\ (3/5)(-2) + (-4/5)(3) \end{pmatrix}
$$

Calculating each component:

First component: (4/5)(-2) = -8/5; (3/5)(3) = 9/5. Adding them: -8/5 + 9/5 = 1/5.

Second component: (3/5)(-2) = -6/5; (-4/5)(3) = -12/5. Adding them: -6/5 -12/5 = -18/5.

So, the result is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$. Let me check if this makes sense.

Alternatively, maybe I can verify if this reflection matrix is correct by checking another vector. For example, let's check if reflecting the original vector $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ gives $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$:

$$
R \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} (4/5)(5) + (3/5)(0) \\ (3/5)(5) + (-4/5)(0) \end{pmatrix} = \begin{pmatrix} 4 + 0 \\ 3 + 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Yes, that checks out. So the matrix is correct.

Alternatively, maybe I can think about the line of reflection. Since the reflection takes (5, 0) to (4, 3), the line of reflection should be the perpendicular bisector of the segment joining (5, 0) and (4, 3). Let me try this approach as a cross-check.

First, find the midpoint between (5, 0) and (4, 3). The midpoint is ((5 + 4)/2, (0 + 3)/2) = (9/2, 3/2). Then, the line of reflection is the perpendicular bisector of the segment connecting these two points.

The slope of the segment connecting (5, 0) and (4, 3) is (3 - 0)/(4 - 5) = 3/(-1) = -3. Therefore, the slope of the perpendicular bisector is the negative reciprocal, which is 1/3.

So, the line of reflection has slope 1/3 and passes through the midpoint (9/2, 3/2). Therefore, its equation is:

y - 3/2 = (1/3)(x - 9/2)

Let me write that in standard form. Multiply both sides by 3:

3y - 9/2 = x - 9/2 ⇒ 3y = x ⇒ x - 3y = 0.

So, the line of reflection is x - 3y = 0, or y = (1/3)x.

Alternatively, if I can find the reflection matrix over the line y = (1/3)x, does it match the matrix I found earlier?

Let me recall that the formula for reflection over a line y = mx is:

$$
R = \frac{1}{1 + m^2} \begin{pmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{pmatrix}
$$

Let me check this formula. If m is the slope, then yes, for a line y = mx, the reflection matrix is as above. Let me verify with m = 1/3.

So, m = 1/3, m² = 1/9.

Then,

R = [1/(1 + 1/9)] * [ [1 - 1/9, 2*(1/3)], [2*(1/3), 1/9 - 1] ]

Compute denominator: 1 + 1/9 = 10/9, so 1/(10/9) = 9/10.

So,

First row: [ (1 - 1/9), 2/3 ] = [8/9, 2/3]

Second row: [2/3, (1/9 - 1)] = [2/3, -8/9]

Multiply each component by 9/10:

First row: (8/9)*(9/10) = 8/10 = 4/5, (2/3)*(9/10) = 6/10 = 3/5

Second row: (2/3)*(9/10) = 6/10 = 3/5, (-8/9)*(9/10) = -8/10 = -4/5

So, the matrix is:

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

Which matches the matrix I found earlier! Great, so that confirms that the reflection is over the line y = (1/3)x, and the matrix is correct.

Therefore, applying this matrix to $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ gives $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$, as calculated before.

But let me just verify once more by another method. For example, using the formula for reflection over a line. Alternatively, maybe using vector projections.

The formula for reflecting a vector v over a line with unit vector u is:

Reflection of v is 2 proj_u(v) - v.

So, if I can find the unit vector in the direction of the line of reflection, which is y = (1/3)x. The direction vector of the line is (3, 1), since slope is 1/3. So, a direction vector is (3, 1). Let me normalize it.

The length of (3, 1) is sqrt(3² + 1²) = sqrt(10). So, unit vector u is (3/sqrt(10), 1/sqrt(10)).

So, for any vector v, its reflection over the line is 2 proj_u(v) - v.

Let me apply this to the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$.

First, compute proj_u(v):

proj_u(v) = (v ⋅ u) u

Compute v ⋅ u:

(-2)(3/sqrt(10)) + 3*(1/sqrt(10)) = (-6 + 3)/sqrt(10) = (-3)/sqrt(10)

Therefore, proj_u(v) = (-3/sqrt(10)) * (3/sqrt(10), 1/sqrt(10)) = (-9/10, -3/10)

Then, reflection is 2*proj_u(v) - v:

2*(-9/10, -3/10) - (-2, 3) = (-18/10, -6/10) + (2, -3) = (-9/5, -3/5) + (2, -3)

Convert 2 to 10/5 and -3 to -15/5:

(-9/5 + 10/5, -3/5 - 15/5) = (1/5, -18/5)

Which is the same result as before! So that confirms that the reflection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ is indeed $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

Therefore, the answer should be $\begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix}$.

But let me check once more with another approach. Let me think about the properties of reflections. For example, if I have a reflection over a line, then the reflection of a point is such that the line is the perpendicular bisector of the segment joining the original point and its image.

So, if I take the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$, let's call it v, and its image under reflection, let's call it w. Then, the midpoint between v and w should lie on the line of reflection, and the vector from v to w should be perpendicular to the line of reflection.

Given that the line of reflection is y = (1/3)x, let me check if this holds for our result.

First, compute the midpoint between v = (-2, 3) and w = (1/5, -18/5). The midpoint is:

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

So, midpoint is (-9/10, -3/10). Let me check if this lies on the line y = (1/3)x.

Compute (1/3)x for x = -9/10: (1/3)(-9/10) = -3/10. Which matches the y-coordinate. So yes, the midpoint lies on the line. Good.

Now, check if the vector from v to w is perpendicular to the line. The vector from v to w is w - v = (1/5 - (-2), -18/5 - 3) = (1/5 + 10/5, -18/5 - 15/5) = (11/5, -33/5). Simplify: (11/5, -33/5) = (11, -33)/5. Which is a scalar multiple of (1, -3). The direction vector of the line is (3, 1), so the direction vector of the line is (3, 1). The vector from v to w is (1, -3). Let me check if they are perpendicular by taking the dot product:

(3, 1) ⋅ (1, -3) = 3*1 + 1*(-3) = 3 - 3 = 0. Yes! They are perpendicular. So that confirms that the reflection is correct.

Therefore, all methods lead to the same conclusion. So, the reflection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

But just to make sure, let me also check if applying the reflection matrix to another vector gives the correct result. For example, take the midpoint we found earlier, (-9/10, -3/10). If I reflect it over the line, it should map to itself. Let me check:

Applying R to (-9/10, -3/10):

First component: (4/5)(-9/10) + (3/5)(-3/10) = (-36/50) + (-9/50) = -45/50 = -9/10

Second component: (3/5)(-9/10) + (-4/5)(-3/10) = (-27/50) + 12/50 = -15/50 = -3/10

Which is the same as the original vector. Perfect, that checks out.

Another check: take a point on the line of reflection, say (3, 1). Reflecting it should give the same point. Let me compute:

R * (3, 1):

First component: (4/5)(3) + (3/5)(1) = 12/5 + 3/5 = 15/5 = 3

Second component: (3/5)(3) + (-4/5)(1) = 9/5 - 4/5 = 5/5 = 1

Yes, so (3, 1) maps to itself. Correct.

Another check: take a point not on the line, say (0, 0). Reflecting it should give itself? Wait, (0, 0) is on the line y = (1/3)x, since 0 = (1/3)*0. So yes, it's on the line. Let me check another point, say (0, 1). Reflecting (0, 1):

R * (0, 1) = (4/5*0 + 3/5*1, 3/5*0 + (-4/5)*1) = (3/5, -4/5). Let me check if this is the reflection over the line y = (1/3)x.

Midpoint between (0, 1) and (3/5, -4/5) is ((0 + 3/5)/2, (1 + (-4/5))/2) = (3/10, (1/5)/2) = (3/10, 1/10). Check if this lies on the line y = (1/3)x: (1/3)(3/10) = 1/10. Yes, matches. Also, the vector from (0, 1) to (3/5, -4/5) is (3/5, -9/5), which is (3, -9)/5, direction vector (3, -9) which is scalar multiple of (1, -3). Which is perpendicular to the line's direction vector (3, 1). Their dot product is 3*1 + 1*(-3) = 0. Correct. So that works.

Therefore, all checks confirm that the reflection matrix is correct, and thus the answer is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

But just to make sure, let me also think if there's another way someone might approach this problem, maybe by using the fact that reflections are involutions (applying them twice gives the identity), but I don't think that's necessary here. Alternatively, maybe using coordinates.

Alternatively, suppose I didn't remember the formula for the reflection matrix. Let me try to find the line of reflection again, and then use the formula for reflection.

We already found that the line of reflection is y = (1/3)x. So, if I can find the reflection of a general point (x, y) over this line, then I can apply it to (-2, 3).

But since I already did that with the projection formula, and it worked, maybe that's redundant. But just to recap, the formula for reflection over y = mx is:

If the line is y = mx, then the reflection of a point (x, y) is given by:

$$
\left( \frac{(1 - m^2)x + 2m y}{1 + m^2}, \frac{2m x - (1 - m^2)y}{1 + m^2} \right)
$$

Let me verify this formula. For m = 1/3, then:

Numerator for x-coordinate: (1 - (1/9))x + 2*(1/3)y = (8/9)x + (2/3)y

Denominator: 1 + 1/9 = 10/9

So, x-coordinate: [ (8/9)x + (2/3)y ] / (10/9) = [ (8x + 6y)/9 ] / (10/9) ) = (8x + 6y)/10 = (4x + 3y)/5

Similarly, y-coordinate: [ 2*(1/3)x - (1 - 1/9)y ] / (10/9) = [ (2/3)x - (8/9)y ] / (10/9) = [ (6x - 8y)/9 ] / (10/9) ) = (6x - 8y)/10 = (3x - 4y)/5

Therefore, the reflection of (x, y) over y = (1/3)x is ( (4x + 3y)/5, (3x - 4y)/5 )

So, applying this to (-2, 3):

x = -2, y = 3

x-coordinate: (4*(-2) + 3*3)/5 = (-8 + 9)/5 = 1/5

y-coordinate: (3*(-2) - 4*3)/5 = (-6 - 12)/5 = -18/5

Which gives the same result! So, this formula also gives the same answer. Therefore, this is another confirmation.

Therefore, regardless of the method, the answer is consistent. So, I can be confident that the reflection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ is $\begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix}$.

Just to recap all steps:

1. Recognize that reflection is a linear transformation, so can be represented by a matrix.
2. Use the given transformation of vector (5, 0) to (4, 3) to set up equations for the matrix.
3. Solve for the matrix entries using the properties of reflection matrices (trace 0, determinant -1).
4. Verify the matrix by checking the determinant and trace, and by applying it to the given vector.
5. Cross-check by finding the line of reflection and using the formula for reflection over a line.
6. Confirm using projection formula and properties of reflections (midpoint on line, perpendicularity).
7. Use the general formula for reflection over a line with slope m to compute the result.
8. All methods lead to the same result, so the answer is confirmed.

Therefore, the final 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 determine where the reflection takes the vector
$$
\begin{pmatrix} -2 \\ 3 \end{pmatrix}
$$
we are given that the same reflection maps
$$
\begin{pmatrix} 5 \\ 0 \end{pmatrix} \mapsto \begin{pmatrix} 4 \\ 3 \end{pmatrix}.
$$

---

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

A reflection in 2D is a **linear transformation** that satisfies the following properties:

- It is **its own inverse**: $ R^2 = I $
- Its **determinant is -1**
- Its **trace is 0**

This allows us to represent the reflection as a 2×2 matrix of the form:
$$
R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
$$
with the constraints:
- $ a + d = 0 $ (trace = 0)
- $ ad - bc = -1 $ (determinant = -1)

---

### Step 2: Use the Given Mapping to Determine the Matrix

We are told:
$$
R \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Apply the matrix multiplication:
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 5a \\ 5c \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

From this, we get:
- $ 5a = 4 \Rightarrow a = \frac{4}{5} $
- $ 5c = 3 \Rightarrow c = \frac{3}{5} $

Using the trace condition $ a + d = 0 $, we find:
- $ d = -a = -\frac{4}{5} $

Now use the determinant condition:
$$
ad - bc = \left( \frac{4}{5} \right)\left( -\frac{4}{5} \right) - b\left( \frac{3}{5} \right) = -\frac{16}{25} - \frac{3b}{5} = -1
$$

Solving for $ b $:
$$
-\frac{16}{25} - \frac{3b}{5} = -1 \Rightarrow \frac{3b}{5} = \frac{9}{25} \Rightarrow b = \frac{3}{5}
$$

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

---

### Step 3: Apply the Matrix to the Target Vector

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

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

---

### Step 4: Final Answer

Thus, the reflection takes the vector
$$
\begin{pmatrix} -2 \\ 3 \end{pmatrix}
$$
to
$$
\boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}
$$

[Assistant]
Explanation:
Okay, so I need to figure out where the reflection takes the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$. The problem says 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. If I can find that matrix, then I can apply it to any vector, including $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$, to find its image under the reflection.

First, let me remember that a reflection matrix in 2D can be represented in different ways depending on the line of reflection. The most common ones are reflections over the x-axis, y-axis, or the line y = x. But since the problem doesn't specify the line, I need to figure out which line is being reflected over based on the given information.

So, given that reflecting $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ gives $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, I can use this information to determine the reflection line. Let me think. If I can find the line of reflection, then I can write the reflection matrix accordingly.

Alternatively, since reflections are linear transformations, maybe I can set up equations to find the reflection matrix. Let me try that approach.

Let me denote the reflection matrix as $ R = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $. Since it's a reflection, it should satisfy certain properties. For example, it should be its own inverse, meaning $ R^2 = I $, and its determinant should be -1. Also, the trace of the matrix should be 0 because for a reflection matrix, trace is 0. Wait, is that true?

Wait, let me check. For a reflection over a line making an angle θ with the x-axis, the reflection matrix is:

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

So, the trace of this matrix is $ \cos 2\theta - \cos 2\theta = 0 $. So yes, the trace is 0. Also, determinant is $ -\cos^2 2\theta - \sin^2 2\theta $? Wait, no. Let me compute determinant:

Determinant of R is $ (a)(d) - (b)(c) $. For the matrix above, that would be $ \cos 2\theta \times (-\cos 2\theta) - \sin 2\theta \times \sin 2\theta = -\cos^2 2\theta - \sin^2 2\theta = -(\cos^2 2\theta + \sin^2 2\theta) = -1 $. So determinant is -1. So, for any reflection matrix, determinant is -1 and trace is 0. So, if I can find such a matrix R that satisfies R * v = w, where v is $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ and w is $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, then I can find the matrix.

So, let me write the equation:

$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Multiplying the matrix with the vector:

First component: 5a + 0*b = 5a = 4 ⇒ a = 4/5

Second component: 5c + 0*d = 5c = 3 ⇒ c = 3/5

So, we have a = 4/5 and c = 3/5. Now, since the matrix is a reflection matrix, we know that trace is 0, so a + d = 0 ⇒ d = -a = -4/5.

Also, determinant is -1, so ad - bc = -1. Let me check that.

We have a = 4/5, d = -4/5, c = 3/5. So, ad - bc = (4/5)(-4/5) - b*(3/5) = -16/25 - (3b)/5 = -1.

So, solving for b:

-16/25 - (3b)/5 = -1 ⇒ - (3b)/5 = -1 + 16/25 = (-25/25 + 16/25) = -9/25 ⇒

Multiply both sides by -1: (3b)/5 = 9/25 ⇒ 3b = 9/5 ⇒ b = 3/5.

So, b = 3/5. Therefore, the reflection matrix is:

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

Let me check if this matrix satisfies the determinant condition. ad - bc = (4/5)(-4/5) - (3/5)(3/5) = -16/25 - 9/25 = -25/25 = -1. Yes, that works. And trace is 4/5 + (-4/5) = 0. Perfect.

So, now that I have the reflection matrix, I can apply it to the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to find its image.

Let me compute that:

$$
R \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \begin{pmatrix} (4/5)(-2) + (3/5)(3) \\ (3/5)(-2) + (-4/5)(3) \end{pmatrix}
$$

Calculating each component:

First component: (4/5)(-2) = -8/5; (3/5)(3) = 9/5. Adding them: -8/5 + 9/5 = 1/5.

Second component: (3/5)(-2) = -6/5; (-4/5)(3) = -12/5. Adding them: -6/5 -12/5 = -18/5.

So, the result is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$. Let me check if this makes sense.

Alternatively, maybe I can verify if this reflection matrix is correct by checking another vector. For example, let's check if reflecting the original vector $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ gives $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$:

$$
R \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} (4/5)(5) + (3/5)(0) \\ (3/5)(5) + (-4/5)(0) \end{pmatrix} = \begin{pmatrix} 4 + 0 \\ 3 + 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Yes, that checks out. So the matrix is correct.

Alternatively, maybe I can think about the line of reflection. Since the reflection takes (5, 0) to (4, 3), the line of reflection should be the perpendicular bisector of the segment joining (5, 0) and (4, 3). Let me try this approach as a cross-check.

First, find the midpoint between (5, 0) and (4, 3). The midpoint is ((5 + 4)/2, (0 + 3)/2) = (9/2, 3/2). Then, the line of reflection is the perpendicular bisector of the segment connecting these two points.

The slope of the segment connecting (5, 0) and (4, 3) is (3 - 0)/(4 - 5) = 3/(-1) = -3. Therefore, the slope of the perpendicular bisector is the negative reciprocal, which is 1/3.

So, the line of reflection has slope 1/3 and passes through the midpoint (9/2, 3/2). Therefore, its equation is:

y - 3/2 = (1/3)(x - 9/2)

Let me write that in standard form. Multiply both sides by 3:

3y - 9/2 = x - 9/2 ⇒ 3y = x ⇒ x - 3y = 0.

So, the line of reflection is x - 3y = 0, or y = (1/3)x.

Alternatively, if I can find the reflection matrix over the line y = (1/3)x, does it match the matrix I found earlier?

Let me recall that the formula for reflection over a line y = mx is:

$$
R = \frac{1}{1 + m^2} \begin{pmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{pmatrix}
$$

Let me check this formula. If m is the slope, then yes, for a line y = mx, the reflection matrix is as above. Let me verify with m = 1/3.

So, m = 1/3, m² = 1/9.

Then,

R = [1/(1 + 1/9)] * [ [1 - 1/9, 2*(1/3)], [2*(1/3), 1/9 - 1] ]

Compute denominator: 1 + 1/9 = 10/9, so 1/(10/9) = 9/10.

So,

First row: [ (1 - 1/9), 2/3 ] = [8/9, 2/3]

Second row: [2/3, (1/9 - 1)] = [2/3, -8/9]

Multiply each component by 9/10:

First row: (8/9)*(9/10) = 8/10 = 4/5, (2/3)*(9/10) = 6/10 = 3/5

Second row: (2/3)*(9/10) = 6/10 = 3/5, (-8/9)*(9/10) = -8/10 = -4/5

So, the matrix is:

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

Which matches the matrix I found earlier! Great, so that confirms that the reflection is over the line y = (1/3)x, and the matrix is correct.

Therefore, applying this matrix to $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ gives $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$, as calculated before.

But let me just verify once more by another method. For example, using the formula for reflection over a line. Alternatively, maybe using vector projections.

The formula for reflecting a vector v over a line with unit vector u is:

Reflection of v is 2 proj_u(v) - v.

So, if I can find the unit vector in the direction of the line of reflection, which is y = (1/3)x. The direction vector of the line is (3, 1), since slope is 1/3. So, a direction vector is (3, 1). Let me normalize it.

The length of (3, 1) is sqrt(3² + 1²) = sqrt(10). So, unit vector u is (3/sqrt(10), 1/sqrt(10)).

So, for any vector v, its reflection over the line is 2 proj_u(v) - v.

Let me apply this to the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$.

First, compute proj_u(v):

proj_u(v) = (v ⋅ u) u

Compute v ⋅ u:

(-2)(3/sqrt(10)) + 3*(1/sqrt(10)) = (-6 + 3)/sqrt(10) = (-3)/sqrt(10)

Therefore, proj_u(v) = (-3/sqrt(10)) * (3/sqrt(10), 1/sqrt(10)) = (-9/10, -3/10)

Then, reflection is 2*proj_u(v) - v:

2*(-9/10, -3/10) - (-2, 3) = (-18/10, -6/10) + (2, -3) = (-9/5, -3/5) + (2, -3)

Convert 2 to 10/5 and -3 to -15/5:

(-9/5 + 10/5, -3/5 - 15/5) = (1/5, -18/5)

Which is the same result as before! So that confirms that the reflection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ is indeed $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

Therefore, the answer should be $\begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix}$.

But let me check once more with another approach. Let me think about the properties of reflections. For example, if I have a reflection over a line, then the reflection of a point is such that the line is the perpendicular bisector of the segment joining the original point and its image.

So, if I take the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$, let's call it v, and its image under reflection, let's call it w. Then, the midpoint between v and w should lie on the line of reflection, and the vector from v to w should be perpendicular to the line of reflection.

Given that the line of reflection is y = (1/3)x, let me check if this holds for our result.

First, compute the midpoint between v = (-2, 3) and w = (1/5, -18/5). The midpoint is:

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

So, midpoint is (-9/10, -3/10). Let me check if this lies on the line y = (1/3)x.

Compute (1/3)x for x = -9/10: (1/3)(-9/10) = -3/10. Which matches the y-coordinate. So yes, the midpoint lies on the line. Good.

Now, check if the vector from v to w is perpendicular to the line. The vector from v to w is w - v = (1/5 - (-2), -18/5 - 3) = (1/5 + 10/5, -18/5 - 15/5) = (11/5, -33/5). Simplify: (11/5, -33/5) = (11, -33)/5. Which is a scalar multiple of (1, -3). The direction vector of the line is (3, 1), so the direction vector of the line is (3, 1). The vector from v to w is (1, -3). Let me check if they are perpendicular by taking the dot product:

(3, 1) ⋅ (1, -3) = 3*1 + 1*(-3) = 3 - 3 = 0. Yes! They are perpendicular. So that confirms that the reflection is correct.

Therefore, all methods lead to the same conclusion. So, the reflection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

But just to make sure, let me also check if applying the reflection matrix to another vector gives the correct result. For example, take the midpoint we found earlier, (-9/10, -3/10). If I reflect it over the line, it should map to itself. Let me check:

Applying R to (-9/10, -3/10):

First component: (4/5)(-9/10) + (3/5)(-3/10) = (-36/50) + (-9/50) = -45/50 = -9/10

Second component: (3/5)(-9/10) + (-4/5)(-3/10) = (-27/50) + 12/50 = -15/50 = -3/10

Which is the same as the original vector. Perfect, that checks out.

Another check: take a point on the line of reflection, say (3, 1). Reflecting it should give the same point. Let me compute:

R * (3, 1):

First component: (4/5)(3) + (3/5)(1) = 12/5 + 3/5 = 15/5 = 3

Second component: (3/5)(3) + (-4/5)(1) = 9/5 - 4/5 = 5/5 = 1

Yes, so (3, 1) maps to itself. Correct.

Another check: take a point not on the line, say (0, 0). Reflecting it should give itself? Wait, (0, 0) is on the line y = (1/3)x, since 0 = (1/3)*0. So yes, it's on the line. Let me check another point, say (0, 1). Reflecting (0, 1):

R * (0, 1) = (4/5*0 + 3/5*1, 3/5*0 + (-4/5)*1) = (3/5, -4/5). Let me check if this is the reflection over the line y = (1/3)x.

Midpoint between (0, 1) and (3/5, -4/5) is ((0 + 3/5)/2, (1 + (-4/5))/2) = (3/10, (1/5)/2) = (3/10, 1/10). Check if this lies on the line y = (1/3)x: (1/3)(3/10) = 1/10. Yes, matches. Also, the vector from (0, 1) to (3/5, -4/5) is (3/5, -9/5), which is (3, -9)/5, direction vector (3, -9) which is scalar multiple of (1, -3). Which is perpendicular to the line's direction vector (3, 1). Their dot product is 3*1 + 1*(-3) = 0. Correct. So that works.

Therefore, all checks confirm that the reflection matrix is correct, and thus the answer is $\begin{pmatrix} 1/5 \\ -18/5 \end{pmatrix}$.

But just to make sure, let me also think if there's another way someone might approach this problem, maybe by using the fact that reflections are involutions (applying them twice gives the identity), but I don't think that's necessary here. Alternatively, maybe using coordinates.

Alternatively, suppose I didn't remember the formula for the reflection matrix. Let me try to find the line of reflection again, and then use the formula for reflection.

We already found that the line of reflection is y = (1/3)x. So, if I can find the reflection of a general point (x, y) over this line, then I can apply it to (-2, 3).

But since I already did that with the projection formula, and it worked, maybe that's redundant. But just to recap, the formula for reflection over y = mx is:

If the line is y = mx, then the reflection of a point (x, y) is given by:

$$
\left( \frac{(1 - m^2)x + 2m y}{1 + m^2}, \frac{2m x - (1 - m^2)y}{1 + m^2} \right)
$$

Let me verify this formula. For m = 1/3, then:

Numerator for x-coordinate: (1 - (1/9))x + 2*(1/3)y = (8/9)x + (2/3)y

Denominator: 1 + 1/9 = 10/9

So, x-coordinate: [ (8/9)x + (2/3)y ] / (10/9) = [ (8x + 6y)/9 ] / (10/9) ) = (8x + 6y)/10 = (4x + 3y)/5

Similarly, y-coordinate: [ 2*(1/3)x - (1 - 1/9)y ] / (10/9) = [ (2/3)x - (8/9)y ] / (10/9) = [ (6x - 8y)/9 ] / (10/9) ) = (6x - 8y)/10 = (3x - 4y)/5

Therefore, the reflection of (x, y) over y = (1/3)x is ( (4x + 3y)/5, (3x - 4y)/5 )

So, applying this to (-2, 3):

x = -2, y = 3

x-coordinate: (4*(-2) + 3*3)/5 = (-8 + 9)/5 = 1/5

y-coordinate: (3*(-2) - 4*3)/5 = (-6 - 12)/5 = -18/5

Which gives the same result! So, this formula also gives the same answer. Therefore, this is another confirmation.

Therefore, regardless of the method, the answer is consistent. So, I can be confident that the reflection of $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ is $\begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix}$.

Just to recap all steps:

1. Recognize that reflection is a linear transformation, so can be represented by a matrix.
2. Use the given transformation of vector (5, 0) to (4, 3) to set up equations for the matrix.
3. Solve for the matrix entries using the properties of reflection matrices (trace 0, determinant -1).
4. Verify the matrix by checking the determinant and trace, and by applying it to the given vector.
5. Cross-check by finding the line of reflection and using the formula for reflection over a line.
6. Confirm using projection formula and properties of reflections (midpoint on line, perpendicularity).
7. Use the general formula for reflection over a line with slope m to compute the result.
8. All methods lead to the same result, so the answer is confirmed.

Therefore, the final 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 determine where the reflection takes the vector
$$
\begin{pmatrix} -2 \\ 3 \end{pmatrix}
$$
we are given that the same reflection maps
$$
\begin{pmatrix} 5 \\ 0 \end{pmatrix} \mapsto \begin{pmatrix} 4 \\ 3 \end{pmatrix}.
$$

---

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

A reflection in 2D is a **linear transformation** that satisfies the following properties:

- It is **its own inverse**: $ R^2 = I $
- Its **determinant is -1**
- Its **trace is 0**

This allows us to represent the reflection as a 2×2 matrix of the form:
$$
R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
$$
with the constraints:
- $ a + d = 0 $ (trace = 0)
- $ ad - bc = -1 $ (determinant = -1)

---

### Step 2: Use the Given Mapping to Determine the Matrix

We are told:
$$
R \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

Apply the matrix multiplication:
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} 5a \\ 5c \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}
$$

From this, we get:
- $ 5a = 4 \Rightarrow a = \frac{4}{5} $
- $ 5c = 3 \Rightarrow c = \frac{3}{5} $

Using the trace condition $ a + d = 0 $, we find:
- $ d = -a = -\frac{4}{5} $

Now use the determinant condition:
$$
ad - bc = \left( \frac{4}{5} \right)\left( -\frac{4}{5} \right) - b\left( \frac{3}{5} \right) = -\frac{16}{25} - \frac{3b}{5} = -1
$$

Solving for $ b $:
$$
-\frac{16}{25} - \frac{3b}{5} = -1 \Rightarrow \frac{3b}{5} = \frac{9}{25} \Rightarrow b = \frac{3}{5}
$$

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

---

### Step 3: Apply the Matrix to the Target Vector

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

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

---

### Step 4: Final Answer

Thus, the reflection takes the vector
$$
\begin{pmatrix} -2 \\ 3 \end{pmatrix}
$$
to
$$
\boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}
$$

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 and maybe Fermat's Little Theorem? Let me recall.

First, maybe I should simplify the bases 129 and 96 modulo 11. That way, instead of dealing with large numbers, I can work with smaller ones. Let me start with 129 divided by 11.

Let me calculate 11 times 11, which is 121. So 129 minus 121 is 8. Therefore, 129 is congruent to 8 modulo 11. So $129 \equiv 8 \mod 11$.

Now for 96. Let me divide 96 by 11. 11 times 8 is 88, so 96 minus 88 is 8. Wait, so 96 is also congruent to 8 modulo 11? Wait, that seems interesting. Let me check again. 11*8=88, 96-88=8. Yes, so 96 ≡ 8 mod 11. So both 129 and 96 are congruent to 8 modulo 11.

So the original expression $129^{34} + 96^{38}$ modulo 11 is equivalent to $8^{34} + 8^{38}$ modulo 11.

So now I need to compute $8^{34} + 8^{38} \mod 11$. Maybe I can factor out some common terms? Let me see.

First, note that $8^{34} + 8^{38} = 8^{34}(1 + 8^{4})$. Because $8^{38} = 8^{34} \times 8^4$. Let me check that exponent: 34 + 4 = 38, yes. So that's correct.

So now, I need to compute $8^{34} \times (1 + 8^4)$ modulo 11.

But maybe there's a better way. Alternatively, since exponents can be simplified using Fermat's Little Theorem. Since 11 is prime, Fermat's Little Theorem says that for any integer a not divisible by 11, $a^{10} \equiv 1 \mod 11$.

So since 8 and 11 are coprime, we can use this theorem. Let me recall that.

So, for example, $8^{10} \equiv 1 \mod 11$. Therefore, exponents can be reduced modulo 10.

So let me compute the exponents 34 and 38 modulo 10.

For 34 divided by 10, the remainder is 4. So 34 = 10*3 + 4. Therefore, $8^{34} \equiv (8^{10})^3 \times 8^4 \equiv 1^3 \times 8^4 \equiv 8^4 \mod 11$.

Similarly, for 38 divided by 10, the remainder is 8. So 38 = 10*3 + 8. Therefore, $8^{38} \equiv (8^{10})^3 \times 8^8 \equiv 1^3 \times 8^8 \equiv 8^8 \mod 11$.

Therefore, the original expression becomes $8^4 + 8^8 \mod 11$.

So now I need to compute $8^4$ and $8^8$ modulo 11. Let me compute these step by step.

First, compute $8^1 \mod 11 = 8$.

$8^2 = 64$. Let me divide 64 by 11. 11*5=55, so 64-55=9. Therefore, $8^2 \equiv 9 \mod 11$.

Then $8^3 = 8^2 * 8 = 9 * 8 = 72$. 72 divided by 11: 11*6=66, so 72-66=6. Therefore, $8^3 \equiv 6 \mod 11$.

$8^4 = 8^3 * 8 = 6 * 8 = 48$. 48 divided by 11: 11*4=44, so 48-44=4. Therefore, $8^4 \equiv 4 \mod 11$.

Okay, so $8^4 \equiv 4$. Now, $8^8$ can be calculated as $(8^4)^2$. Since $8^4 \equiv 4$, then $8^8 \equiv 4^2 = 16 \mod 11$. 16 divided by 11 is 1 with remainder 5, so $8^8 \equiv 5 \mod 11$.

Therefore, the original expression $8^4 + 8^8 \equiv 4 + 5 = 9 \mod 11$.

Wait, so the remainder is 9? Let me check my steps again to make sure I didn't make a mistake.

First, simplifying 129 and 96 modulo 11:

129 divided by 11: 11*11=121, 129-121=8. Correct.

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

So both are 8 mod 11.

Then, $8^{34} + 8^{38}$.

Using Fermat's Little Theorem: since 8^10 ≡ 1 mod 11.

So 34 divided by 10: 34 = 3*10 + 4, so 8^34 ≡ 8^4 mod 11.

Similarly, 38 = 3*10 + 8, so 8^38 ≡ 8^8 mod 11.

Then, computing 8^4:

8^1=8, 8^2=64≡9, 8^3=8*9=72≡6, 8^4=8*6=48≡4. Correct.

Then 8^8 is (8^4)^2 ≡ 4^2=16≡5. Correct.

So 4 + 5 = 9. Therefore, the remainder is 9.

Wait, but let me check another way. Maybe compute 8^4 + 8^8 mod 11. Alternatively, maybe compute 8^4 + 8^8 as 8^4(1 + 8^4). Wait, earlier I thought of factoring, but maybe that's another way.

So 8^4(1 + 8^4) mod 11.

We already found 8^4 ≡ 4 mod 11. Then 1 + 8^4 ≡ 1 + 4 = 5 mod 11. Therefore, 8^4 * 5 ≡ 4 * 5 = 20 ≡ 20 - 11 = 9 mod 11. Same result.

So that's consistent.

Alternatively, maybe compute 8^4 + 8^8 directly.

But 8^4 is 4, 8^8 is 5, so 4 + 5 = 9.

Alternatively, maybe compute 8^4 + 8^8 mod 11.

Alternatively, maybe compute exponents step by step. Let me check 8^4 again.

8^1 = 8 mod 11.

8^2 = 64 mod 11. 64 / 11 is 5*11=55, 64-55=9. So 9.

8^3 = 8^2 * 8 = 9*8=72. 72 / 11 is 6*11=66, 72-66=6.

8^4 = 6*8=48. 48-44=4.

8^5 = 4*8=32. 32-22=10.

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

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

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

8^8 = 2*8=16. 16 mod 11 is 5.

So that's the same as before. So 8^8 ≡ 5.

So 8^4 + 8^8 = 4 + 5 = 9.

Therefore, the remainder is 9.

But let me check once more with another approach. Maybe using exponents in another way.

Alternatively, since 8 ≡ -3 mod 11. Because 8 + 3 = 11, so 8 ≡ -3 mod 11. Maybe that's easier? Let me try.

So 8 ≡ -3 mod 11. Therefore, 8^k ≡ (-3)^k mod 11.

So let's compute 8^34 + 8^38 ≡ (-3)^{34} + (-3)^{38} mod 11.

Note that (-3)^{34} = [(-3)^2]^{17} = 9^{17} mod 11. Similarly, (-3)^{38} = [(-3)^2]^{19} = 9^{19} mod 11.

But maybe that's more complicated? Let me see. Alternatively, since (-3)^k = (-1)^k * 3^k.

So (-3)^{34} = (-1)^{34} * 3^{34} = 1 * 3^{34} mod 11. Similarly, (-3)^{38} = (-1)^{38} * 3^{38} = 1 * 3^{38} mod 11.

Therefore, the expression becomes 3^{34} + 3^{38} mod 11.

Hmm, maybe this is another way. Let me see if I can compute this.

But maybe using Fermat's theorem again. Since 3^10 ≡ 1 mod 11.

So 3^{34} = 3^{10*3 + 4} = (3^{10})^3 * 3^4 ≡ 1^3 * 3^4 ≡ 3^4 mod 11.

Similarly, 3^{38} = 3^{10*3 + 8} = (3^{10})^3 * 3^8 ≡ 1^3 * 3^8 ≡ 3^8 mod 11.

Therefore, the expression is 3^4 + 3^8 mod 11.

Let me compute 3^4 and 3^8.

3^1 = 3 mod 11.

3^2 = 9 mod 11.

3^3 = 27 mod 11. 27 - 22 = 5. So 5.

3^4 = 5 * 3 = 15 mod 11. 15 - 11 = 4. So 4.

3^4 ≡ 4 mod 11.

Then 3^8 = (3^4)^2 ≡ 4^2 = 16 ≡ 5 mod 11.

Therefore, 3^4 + 3^8 ≡ 4 + 5 = 9 mod 11.

Same result! So that's consistent.

Therefore, regardless of the method, I get 9.

Alternatively, maybe I can check with another approach. Let me try calculating 8^4 + 8^8 mod 11 again.

We had 8^4 ≡ 4, 8^8 ≡ 5. So 4 + 5 = 9.

Alternatively, maybe compute 8^4 + 8^8 mod 11 as follows:

Note that 8^4 ≡ 4, so 8^4 + 8^8 = 4 + (8^4)^2 = 4 + 4^2 = 4 + 16 = 20 ≡ 20 - 11 = 9 mod 11.

Same result.

So all methods lead to 9. Therefore, I think the remainder is 9.

But just to be thorough, let me check with actual exponentiation.

Alternatively, maybe compute 8^4 mod 11: 8^4 = 4096. Let me divide 4096 by 11.

But that's tedious. Alternatively, since we already did it step by step and got 4, which is correct.

Alternatively, 8^4 = (8^2)^2 = 64^2. 64 mod 11 is 9, so 9^2 = 81 mod 11. 81 divided by 11 is 7*11=77, 81-77=4. So 4. Correct.

Similarly, 8^8 = (8^4)^2 ≡ 4^2 = 16 ≡ 5. Correct.

So 4 + 5 = 9.

Therefore, I think the answer is 9.

But just to make sure, let me check with another exponent reduction.

Alternatively, since 8^10 ≡ 1 mod 11, then 8^34 = 8^(10*3 + 4) = (8^10)^3 * 8^4 ≡ 1^3 * 8^4 ≡ 8^4. Similarly, 8^38 = 8^(10*3 + 8) ≡ 8^8.

So again, same as before.

Alternatively, maybe compute 8^4 + 8^8 mod 11 as follows:

Let me compute 8^4 + 8^8 = 8^4(1 + 8^4) ≡ 4*(1 + 4) = 4*5 = 20 ≡ 9 mod 11.

Same result.

Therefore, I think I can be confident that the remainder is 9.

But just to be 100% sure, let me check with smaller exponents.

For example, let me compute 8^1 + 8^1 = 16 mod 11 is 5.

But if I use the same method: 8^1 + 8^1 = 8 + 8 = 16 ≡ 5 mod 11.

Alternatively, if I use exponents: 8^1 + 8^1 = 8^1*(1 + 8^0) = 8*(1 + 1) = 16 ≡ 5.

But that's a different exponent.

Alternatively, maybe check with exponents that I can compute manually.

For example, compute 8^2 + 8^3 mod 11.

Which is 64 + 512. 64 mod 11 is 9, 512 mod 11. Let me compute 512 / 11.

Well, 11*46 = 506, so 512 - 506 = 6. Therefore, 8^3 ≡ 6 mod 11. Therefore, 9 + 6 = 15 ≡ 4 mod 11.

Alternatively, using our method: 8^2 + 8^3 = 8^2(1 + 8^1) = 9*(1 + 8) = 9*9 = 81 ≡ 4 mod 11. Which matches.

So that works.

Therefore, the method is correct.

Therefore, I think my answer is correct.

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

To find the remainder when $129^{34} + 96^{38}$ is divided by 11, we can simplify the problem using **modular arithmetic** and **Fermat's Little Theorem**.

---

### Step 1: Reduce the Bases Modulo 11

We begin by reducing the bases modulo 11:

- $129 \mod 11$:
  $11 \times 11 = 121$, so $129 - 121 = 8$
  Therefore, $129 \equiv 8 \mod 11$

- $96 \mod 11$:
  $11 \times 8 = 88$, so $96 - 88 = 8$
  Therefore, $96 \equiv 8 \mod 11$

So, the original expression becomes:
$$
129^{34} + 96^{38} \equiv 8^{34} + 8^{38} \mod 11
$$

---

### Step 2: Apply Fermat's Little Theorem

Since 11 is a prime number, **Fermat's Little Theorem** tells us that for any integer $a$ not divisible by 11:
$$
a^{10} \equiv 1 \mod 11
$$

We can use this to reduce the exponents modulo 10:

- $34 \mod 10 = 4$, so $8^{34} \equiv 8^4 \mod 11$
- $38 \mod 10 = 8$, so $8^{38} \equiv 8^8 \mod 11$

Thus, the expression becomes:
$$
8^{34} + 8^{38} \equiv 8^4 + 8^8 \mod 11
$$

---

### Step 3: Compute Powers of 8 Modulo 11

We compute the powers of 8 modulo 11 step by step:

- $8^1 \equiv 8 \mod 11$
- $8^2 = 64 \equiv 9 \mod 11$
- $8^3 = 8 \cdot 9 = 72 \equiv 6 \mod 11$
- $8^4 = 8 \cdot 6 = 48 \equiv 4 \mod 11$
- $8^8 = (8^4)^2 = 4^2 = 16 \equiv 5 \mod 11$

So:
$$
8^4 \equiv 4 \mod 11, \quad 8^8 \equiv 5 \mod 11
$$

---

### Step 4: Add the Results

Now, we add the two results:
$$
8^4 + 8^8 \equiv 4 + 5 = 9 \mod 11
$$

---

### 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 and maybe Fermat's Little Theorem? Let me recall.

First, maybe I should simplify the bases 129 and 96 modulo 11. That way, instead of dealing with large numbers, I can work with smaller ones. Let me start with 129 divided by 11.

Let me calculate 11 times 11, which is 121. So 129 minus 121 is 8. Therefore, 129 is congruent to 8 modulo 11. So $129 \equiv 8 \mod 11$.

Now for 96. Let me divide 96 by 11. 11 times 8 is 88, so 96 minus 88 is 8. Wait, so 96 is also congruent to 8 modulo 11? Wait, that seems interesting. Let me check again. 11*8=88, 96-88=8. Yes, so 96 ≡ 8 mod 11. So both 129 and 96 are congruent to 8 modulo 11.

So the original expression $129^{34} + 96^{38}$ modulo 11 is equivalent to $8^{34} + 8^{38}$ modulo 11.

So now I need to compute $8^{34} + 8^{38} \mod 11$. Maybe I can factor out some common terms? Let me see.

First, note that $8^{34} + 8^{38} = 8^{34}(1 + 8^{4})$. Because $8^{38} = 8^{34} \times 8^4$. Let me check that exponent: 34 + 4 = 38, yes. So that's correct.

So now, I need to compute $8^{34} \times (1 + 8^4)$ modulo 11.

But maybe there's a better way. Alternatively, since exponents can be simplified using Fermat's Little Theorem. Since 11 is prime, Fermat's Little Theorem says that for any integer a not divisible by 11, $a^{10} \equiv 1 \mod 11$.

So since 8 and 11 are coprime, we can use this theorem. Let me recall that.

So, for example, $8^{10} \equiv 1 \mod 11$. Therefore, exponents can be reduced modulo 10.

So let me compute the exponents 34 and 38 modulo 10.

For 34 divided by 10, the remainder is 4. So 34 = 10*3 + 4. Therefore, $8^{34} \equiv (8^{10})^3 \times 8^4 \equiv 1^3 \times 8^4 \equiv 8^4 \mod 11$.

Similarly, for 38 divided by 10, the remainder is 8. So 38 = 10*3 + 8. Therefore, $8^{38} \equiv (8^{10})^3 \times 8^8 \equiv 1^3 \times 8^8 \equiv 8^8 \mod 11$.

Therefore, the original expression becomes $8^4 + 8^8 \mod 11$.

So now I need to compute $8^4$ and $8^8$ modulo 11. Let me compute these step by step.

First, compute $8^1 \mod 11 = 8$.

$8^2 = 64$. Let me divide 64 by 11. 11*5=55, so 64-55=9. Therefore, $8^2 \equiv 9 \mod 11$.

Then $8^3 = 8^2 * 8 = 9 * 8 = 72$. 72 divided by 11: 11*6=66, so 72-66=6. Therefore, $8^3 \equiv 6 \mod 11$.

$8^4 = 8^3 * 8 = 6 * 8 = 48$. 48 divided by 11: 11*4=44, so 48-44=4. Therefore, $8^4 \equiv 4 \mod 11$.

Okay, so $8^4 \equiv 4$. Now, $8^8$ can be calculated as $(8^4)^2$. Since $8^4 \equiv 4$, then $8^8 \equiv 4^2 = 16 \mod 11$. 16 divided by 11 is 1 with remainder 5, so $8^8 \equiv 5 \mod 11$.

Therefore, the original expression $8^4 + 8^8 \equiv 4 + 5 = 9 \mod 11$.

Wait, so the remainder is 9? Let me check my steps again to make sure I didn't make a mistake.

First, simplifying 129 and 96 modulo 11:

129 divided by 11: 11*11=121, 129-121=8. Correct.

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

So both are 8 mod 11.

Then, $8^{34} + 8^{38}$.

Using Fermat's Little Theorem: since 8^10 ≡ 1 mod 11.

So 34 divided by 10: 34 = 3*10 + 4, so 8^34 ≡ 8^4 mod 11.

Similarly, 38 = 3*10 + 8, so 8^38 ≡ 8^8 mod 11.

Then, computing 8^4:

8^1=8, 8^2=64≡9, 8^3=8*9=72≡6, 8^4=8*6=48≡4. Correct.

Then 8^8 is (8^4)^2 ≡ 4^2=16≡5. Correct.

So 4 + 5 = 9. Therefore, the remainder is 9.

Wait, but let me check another way. Maybe compute 8^4 + 8^8 mod 11. Alternatively, maybe compute 8^4 + 8^8 as 8^4(1 + 8^4). Wait, earlier I thought of factoring, but maybe that's another way.

So 8^4(1 + 8^4) mod 11.

We already found 8^4 ≡ 4 mod 11. Then 1 + 8^4 ≡ 1 + 4 = 5 mod 11. Therefore, 8^4 * 5 ≡ 4 * 5 = 20 ≡ 20 - 11 = 9 mod 11. Same result.

So that's consistent.

Alternatively, maybe compute 8^4 + 8^8 directly.

But 8^4 is 4, 8^8 is 5, so 4 + 5 = 9.

Alternatively, maybe compute 8^4 + 8^8 mod 11.

Alternatively, maybe compute exponents step by step. Let me check 8^4 again.

8^1 = 8 mod 11.

8^2 = 64 mod 11. 64 / 11 is 5*11=55, 64-55=9. So 9.

8^3 = 8^2 * 8 = 9*8=72. 72 / 11 is 6*11=66, 72-66=6.

8^4 = 6*8=48. 48-44=4.

8^5 = 4*8=32. 32-22=10.

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

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

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

8^8 = 2*8=16. 16 mod 11 is 5.

So that's the same as before. So 8^8 ≡ 5.

So 8^4 + 8^8 = 4 + 5 = 9.

Therefore, the remainder is 9.

But let me check once more with another approach. Maybe using exponents in another way.

Alternatively, since 8 ≡ -3 mod 11. Because 8 + 3 = 11, so 8 ≡ -3 mod 11. Maybe that's easier? Let me try.

So 8 ≡ -3 mod 11. Therefore, 8^k ≡ (-3)^k mod 11.

So let's compute 8^34 + 8^38 ≡ (-3)^{34} + (-3)^{38} mod 11.

Note that (-3)^{34} = [(-3)^2]^{17} = 9^{17} mod 11. Similarly, (-3)^{38} = [(-3)^2]^{19} = 9^{19} mod 11.

But maybe that's more complicated? Let me see. Alternatively, since (-3)^k = (-1)^k * 3^k.

So (-3)^{34} = (-1)^{34} * 3^{34} = 1 * 3^{34} mod 11. Similarly, (-3)^{38} = (-1)^{38} * 3^{38} = 1 * 3^{38} mod 11.

Therefore, the expression becomes 3^{34} + 3^{38} mod 11.

Hmm, maybe this is another way. Let me see if I can compute this.

But maybe using Fermat's theorem again. Since 3^10 ≡ 1 mod 11.

So 3^{34} = 3^{10*3 + 4} = (3^{10})^3 * 3^4 ≡ 1^3 * 3^4 ≡ 3^4 mod 11.

Similarly, 3^{38} = 3^{10*3 + 8} = (3^{10})^3 * 3^8 ≡ 1^3 * 3^8 ≡ 3^8 mod 11.

Therefore, the expression is 3^4 + 3^8 mod 11.

Let me compute 3^4 and 3^8.

3^1 = 3 mod 11.

3^2 = 9 mod 11.

3^3 = 27 mod 11. 27 - 22 = 5. So 5.

3^4 = 5 * 3 = 15 mod 11. 15 - 11 = 4. So 4.

3^4 ≡ 4 mod 11.

Then 3^8 = (3^4)^2 ≡ 4^2 = 16 ≡ 5 mod 11.

Therefore, 3^4 + 3^8 ≡ 4 + 5 = 9 mod 11.

Same result! So that's consistent.

Therefore, regardless of the method, I get 9.

Alternatively, maybe I can check with another approach. Let me try calculating 8^4 + 8^8 mod 11 again.

We had 8^4 ≡ 4, 8^8 ≡ 5. So 4 + 5 = 9.

Alternatively, maybe compute 8^4 + 8^8 mod 11 as follows:

Note that 8^4 ≡ 4, so 8^4 + 8^8 = 4 + (8^4)^2 = 4 + 4^2 = 4 + 16 = 20 ≡ 20 - 11 = 9 mod 11.

Same result.

So all methods lead to 9. Therefore, I think the remainder is 9.

But just to be thorough, let me check with actual exponentiation.

Alternatively, maybe compute 8^4 mod 11: 8^4 = 4096. Let me divide 4096 by 11.

But that's tedious. Alternatively, since we already did it step by step and got 4, which is correct.

Alternatively, 8^4 = (8^2)^2 = 64^2. 64 mod 11 is 9, so 9^2 = 81 mod 11. 81 divided by 11 is 7*11=77, 81-77=4. So 4. Correct.

Similarly, 8^8 = (8^4)^2 ≡ 4^2 = 16 ≡ 5. Correct.

So 4 + 5 = 9.

Therefore, I think the answer is 9.

But just to make sure, let me check with another exponent reduction.

Alternatively, since 8^10 ≡ 1 mod 11, then 8^34 = 8^(10*3 + 4) = (8^10)^3 * 8^4 ≡ 1^3 * 8^4 ≡ 8^4. Similarly, 8^38 = 8^(10*3 + 8) ≡ 8^8.

So again, same as before.

Alternatively, maybe compute 8^4 + 8^8 mod 11 as follows:

Let me compute 8^4 + 8^8 = 8^4(1 + 8^4) ≡ 4*(1 + 4) = 4*5 = 20 ≡ 9 mod 11.

Same result.

Therefore, I think I can be confident that the remainder is 9.

But just to be 100% sure, let me check with smaller exponents.

For example, let me compute 8^1 + 8^1 = 16 mod 11 is 5.

But if I use the same method: 8^1 + 8^1 = 8 + 8 = 16 ≡ 5 mod 11.

Alternatively, if I use exponents: 8^1 + 8^1 = 8^1*(1 + 8^0) = 8*(1 + 1) = 16 ≡ 5.

But that's a different exponent.

Alternatively, maybe check with exponents that I can compute manually.

For example, compute 8^2 + 8^3 mod 11.

Which is 64 + 512. 64 mod 11 is 9, 512 mod 11. Let me compute 512 / 11.

Well, 11*46 = 506, so 512 - 506 = 6. Therefore, 8^3 ≡ 6 mod 11. Therefore, 9 + 6 = 15 ≡ 4 mod 11.

Alternatively, using our method: 8^2 + 8^3 = 8^2(1 + 8^1) = 9*(1 + 8) = 9*9 = 81 ≡ 4 mod 11. Which matches.

So that works.

Therefore, the method is correct.

Therefore, I think my answer is correct.

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

To find the remainder when $129^{34} + 96^{38}$ is divided by 11, we can simplify the problem using **modular arithmetic** and **Fermat's Little Theorem**.

---

### Step 1: Reduce the Bases Modulo 11

We begin by reducing the bases modulo 11:

- $129 \mod 11$:
  $11 \times 11 = 121$, so $129 - 121 = 8$
  Therefore, $129 \equiv 8 \mod 11$

- $96 \mod 11$:
  $11 \times 8 = 88$, so $96 - 88 = 8$
  Therefore, $96 \equiv 8 \mod 11$

So, the original expression becomes:
$$
129^{34} + 96^{38} \equiv 8^{34} + 8^{38} \mod 11
$$

---

### Step 2: Apply Fermat's Little Theorem

Since 11 is a prime number, **Fermat's Little Theorem** tells us that for any integer $a$ not divisible by 11:
$$
a^{10} \equiv 1 \mod 11
$$

We can use this to reduce the exponents modulo 10:

- $34 \mod 10 = 4$, so $8^{34} \equiv 8^4 \mod 11$
- $38 \mod 10 = 8$, so $8^{38} \equiv 8^8 \mod 11$

Thus, the expression becomes:
$$
8^{34} + 8^{38} \equiv 8^4 + 8^8 \mod 11
$$

---

### Step 3: Compute Powers of 8 Modulo 11

We compute the powers of 8 modulo 11 step by step:

- $8^1 \equiv 8 \mod 11$
- $8^2 = 64 \equiv 9 \mod 11$
- $8^3 = 8 \cdot 9 = 72 \equiv 6 \mod 11$
- $8^4 = 8 \cdot 6 = 48 \equiv 4 \mod 11$
- $8^8 = (8^4)^2 = 4^2 = 16 \equiv 5 \mod 11$

So:
$$
8^4 \equiv 4 \mod 11, \quad 8^8 \equiv 5 \mod 11
$$

---

### Step 4: Add the Results

Now, we add the two results:
$$
8^4 + 8^8 \equiv 4 + 5 = 9 \mod 11
$$

---

### 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 ax² + bx + c, the sum of the coefficients is a + b + c, which is the same as plugging in x = 1: a(1)² + b(1) + c = a + b + c. So, applying that here, the sum of the coefficients of g(x) would be g(1).

But wait, g(x) is f(x - 1). So, g(1) would be f(1 - 1) = f(0). Therefore, the sum of the coefficients of g(x) is f(0). Is that right? Let me verify.

Yes, because if you substitute x = 1 into g(x) = f(x - 1), you get f(0). And since the sum of the coefficients of g(x) is g(1), which equals f(0). Therefore, I just need to compute f(0) and that will be the answer.

But let me make sure I didn't skip any steps or make a mistake here. Let me recall: For any polynomial h(x), the sum of its coefficients is h(1). So if g(x) = f(x - 1), then sum of coefficients of g is g(1) = f(1 - 1) = f(0). Therefore, yes, f(0) is the answer.

So, 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 sum of the coefficients of g(x) equal to -2? That seems straightforward, but let me check if there's another way to approach this to confirm.

Alternatively, maybe I can expand f(x - 1) and then find the sum of its coefficients. Let me try that method as a check.

First, let's write f(x) = 3x⁴ + 5x² - 9x - 2. Then, f(x - 1) would be substituting (x - 1) into each term. Let me compute each term step by step.

First term: 3(x - 1)^4. Let me expand that.

(x - 1)^4 can be expanded using the binomial theorem. Let me recall that (a - b)^4 = a^4 - 4a³b + 6a²b² - 4ab³ + b^4. So here, a = x, b = 1. Therefore:

(x - 1)^4 = x^4 - 4x³ + 6x² - 4x + 1. Multiply by 3:

3x^4 - 12x³ + 18x² - 12x + 3.

Second term: 5(x - 1)^2. Let me expand that. (x - 1)^2 = x² - 2x + 1. Multiply by 5:

5x² - 10x + 5.

Third term: -9(x - 1). That's straightforward:

-9x + 9.

Fourth term: -2. That's just -2.

Now, let me add all these terms together:

First term: 3x^4 - 12x³ + 18x² - 12x + 3

Second term: + 5x² - 10x + 5

Third term: -9x + 9

Fourth term: -2

Now, combine like terms.

Start with the x^4 term: 3x^4.

Next, x³ term: -12x³.

x² terms: 18x² + 5x² = 23x².

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

Constant terms: 3 + 5 + 9 - 2 = (3 + 5) + (9 - 2) = 8 + 7 = 15.

So putting it all together, g(x) = 3x^4 - 12x³ + 23x² - 31x + 15.

Now, the sum of the coefficients is obtained by plugging in x = 1:

g(1) = 3(1)^4 - 12(1)^3 + 23(1)^2 - 31(1) + 15 = 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, that gives -2, which matches the previous result. Therefore, the sum of the coefficients of g(x) is indeed -2.

Therefore, both methods confirm that the answer is -2.

But just to make sure I didn't make any mistakes in expanding f(x - 1), let me check the expansion again. Let me verify each term:

First term: 3(x - 1)^4. As above, that's 3(x^4 - 4x³ + 6x² - 4x + 1) = 3x^4 - 12x³ + 18x² - 12x + 3. Correct.

Second term: 5(x - 1)^2. (x - 1)^2 is x² - 2x + 1, multiplied by 5 gives 5x² - 10x + 5. Correct.

Third term: -9(x - 1) = -9x + 9. Correct.

Fourth term: -2. Correct.

Adding them up:

3x^4 -12x³ +18x² -12x +3 +5x² -10x +5 -9x +9 -2.

Combine x^4: 3x^4.

x³: -12x³.

x²: 18x² +5x² = 23x².

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

Constants: 3 +5 +9 -2 = 15. Correct.

So, the expansion is correct. Then evaluating at x=1 gives -2.

Alternatively, if I didn't want to expand the entire polynomial, I could have just evaluated f(0) as I first thought. Let me check that again.

Original f(x) = 3x⁴ +5x² -9x -2. So f(0) is 0 + 0 -0 -2 = -2. So that's straightforward.

Therefore, both methods confirm that the sum of the coefficients of g(x) is -2.

But just to make sure that my initial reasoning was correct, let me think again. The sum of coefficients of g(x) is g(1). Since g(x) = f(x - 1), then g(1) = f(1 - 1) = f(0). Therefore, the sum is f(0) = -2.

Yes, that's correct. So regardless of the method, the answer is -2.

Therefore, I can be confident that the answer is -2.

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

To find the sum of the coefficients of the polynomial $ g(x) = f(x - 1) $, we begin by recalling a key property of polynomials:

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

---

### Step 1: Understand the relationship between $ f(x) $ and $ g(x) $

We are given:
$$
f(x) = 3x^4 + 5x^2 - 9x - 2
$$
and
$$
g(x) = f(x - 1)
$$

So, $ g(x) $ is the result of substituting $ x - 1 $ into $ f(x) $.

---

### Step 2: Use the property of sum of coefficients

As mentioned, the sum of the coefficients of $ g(x) $ is:
$$
g(1) = f(1 - 1) = f(0)
$$

So, instead of expanding $ f(x - 1) $, we can directly compute $ f(0) $.

---

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

$$
f(0) = 3(0)^4 + 5(0)^2 - 9(0) - 2 = 0 + 0 - 0 - 2 = -2
$$

---

### Step 4: Final Answer

Thus, 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 ax² + bx + c, the sum of the coefficients is a + b + c, which is the same as plugging in x = 1: a(1)² + b(1) + c = a + b + c. So, applying that here, the sum of the coefficients of g(x) would be g(1).

But wait, g(x) is f(x - 1). So, g(1) would be f(1 - 1) = f(0). Therefore, the sum of the coefficients of g(x) is f(0). Is that right? Let me verify.

Yes, because if you substitute x = 1 into g(x) = f(x - 1), you get f(0). And since the sum of the coefficients of g(x) is g(1), which equals f(0). Therefore, I just need to compute f(0) and that will be the answer.

But let me make sure I didn't skip any steps or make a mistake here. Let me recall: For any polynomial h(x), the sum of its coefficients is h(1). So if g(x) = f(x - 1), then sum of coefficients of g is g(1) = f(1 - 1) = f(0). Therefore, yes, f(0) is the answer.

So, 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 sum of the coefficients of g(x) equal to -2? That seems straightforward, but let me check if there's another way to approach this to confirm.

Alternatively, maybe I can expand f(x - 1) and then find the sum of its coefficients. Let me try that method as a check.

First, let's write f(x) = 3x⁴ + 5x² - 9x - 2. Then, f(x - 1) would be substituting (x - 1) into each term. Let me compute each term step by step.

First term: 3(x - 1)^4. Let me expand that.

(x - 1)^4 can be expanded using the binomial theorem. Let me recall that (a - b)^4 = a^4 - 4a³b + 6a²b² - 4ab³ + b^4. So here, a = x, b = 1. Therefore:

(x - 1)^4 = x^4 - 4x³ + 6x² - 4x + 1. Multiply by 3:

3x^4 - 12x³ + 18x² - 12x + 3.

Second term: 5(x - 1)^2. Let me expand that. (x - 1)^2 = x² - 2x + 1. Multiply by 5:

5x² - 10x + 5.

Third term: -9(x - 1). That's straightforward:

-9x + 9.

Fourth term: -2. That's just -2.

Now, let me add all these terms together:

First term: 3x^4 - 12x³ + 18x² - 12x + 3

Second term: + 5x² - 10x + 5

Third term: -9x + 9

Fourth term: -2

Now, combine like terms.

Start with the x^4 term: 3x^4.

Next, x³ term: -12x³.

x² terms: 18x² + 5x² = 23x².

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

Constant terms: 3 + 5 + 9 - 2 = (3 + 5) + (9 - 2) = 8 + 7 = 15.

So putting it all together, g(x) = 3x^4 - 12x³ + 23x² - 31x + 15.

Now, the sum of the coefficients is obtained by plugging in x = 1:

g(1) = 3(1)^4 - 12(1)^3 + 23(1)^2 - 31(1) + 15 = 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, that gives -2, which matches the previous result. Therefore, the sum of the coefficients of g(x) is indeed -2.

Therefore, both methods confirm that the answer is -2.

But just to make sure I didn't make any mistakes in expanding f(x - 1), let me check the expansion again. Let me verify each term:

First term: 3(x - 1)^4. As above, that's 3(x^4 - 4x³ + 6x² - 4x + 1) = 3x^4 - 12x³ + 18x² - 12x + 3. Correct.

Second term: 5(x - 1)^2. (x - 1)^2 is x² - 2x + 1, multiplied by 5 gives 5x² - 10x + 5. Correct.

Third term: -9(x - 1) = -9x + 9. Correct.

Fourth term: -2. Correct.

Adding them up:

3x^4 -12x³ +18x² -12x +3 +5x² -10x +5 -9x +9 -2.

Combine x^4: 3x^4.

x³: -12x³.

x²: 18x² +5x² = 23x².

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

Constants: 3 +5 +9 -2 = 15. Correct.

So, the expansion is correct. Then evaluating at x=1 gives -2.

Alternatively, if I didn't want to expand the entire polynomial, I could have just evaluated f(0) as I first thought. Let me check that again.

Original f(x) = 3x⁴ +5x² -9x -2. So f(0) is 0 + 0 -0 -2 = -2. So that's straightforward.

Therefore, both methods confirm that the sum of the coefficients of g(x) is -2.

But just to make sure that my initial reasoning was correct, let me think again. The sum of coefficients of g(x) is g(1). Since g(x) = f(x - 1), then g(1) = f(1 - 1) = f(0). Therefore, the sum is f(0) = -2.

Yes, that's correct. So regardless of the method, the answer is -2.

Therefore, I can be confident that the answer is -2.

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

To find the sum of the coefficients of the polynomial $ g(x) = f(x - 1) $, we begin by recalling a key property of polynomials:

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

---

### Step 1: Understand the relationship between $ f(x) $ and $ g(x) $

We are given:
$$
f(x) = 3x^4 + 5x^2 - 9x - 2
$$
and
$$
g(x) = f(x - 1)
$$

So, $ g(x) $ is the result of substituting $ x - 1 $ into $ f(x) $.

---

### Step 2: Use the property of sum of coefficients

As mentioned, the sum of the coefficients of $ g(x) $ is:
$$
g(1) = f(1 - 1) = f(0)
$$

So, instead of expanding $ f(x - 1) $, we can directly compute $ f(0) $.

---

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

$$
f(0) = 3(0)^4 + 5(0)^2 - 9(0) - 2 = 0 + 0 - 0 - 2 = -2
$$

---

### Step 4: Final Answer

Thus, 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 that the shortest distance from a point to a line is the perpendicular distance. So maybe I can use the formula for the distance from a point to a line? Wait, what's that formula again?

I think it's something like |Ax + By + C| divided by the square root of A squared plus B squared. But let me make sure. The general formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / sqrt(A² + B²). Right, that sounds familiar. So first, I need to write the given line equation in the standard form Ax + By + C = 0.

The given line is y = 2x - 2. Let me rearrange that. Subtract 2x from both sides: -2x + y + 2 = 0. Or maybe I can write it as 2x - y - 2 = 0. Wait, let me check. If I move all terms to the left side: y - 2x + 2 = 0? Wait, no. Let me do 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. So that's the standard form. Alternatively, multiplying both sides by -1 to make the coefficient of x positive: 2x - y - 2 = 0. Either way, both forms are acceptable. Let me pick one. Let me go with 2x - y - 2 = 0. So A is 2, B is -1, and C is -2.

Wait, but in the formula, it's Ax + By + C = 0. So if I have 2x - y - 2 = 0, then A = 2, B = -1, C = -2. So then plugging into the distance formula: |A*x₀ + B*y₀ + C| / sqrt(A² + B²). Let me check that again. So the point is (6, 0). So x₀ is 6, y₀ is 0.

So substituting into the numerator: |2*6 + (-1)*0 + (-2)|. Let me compute that. 2*6 is 12, (-1)*0 is 0, and then -2. So 12 + 0 - 2 = 10. So absolute value of 10 is 10. Then the denominator is sqrt(2² + (-1)^2) which is sqrt(4 + 1) = sqrt(5). Therefore, the distance is 10 / sqrt(5). But they want it in simplest radical form. So rationalizing the denominator: 10 / sqrt(5) is equal to (10*sqrt(5)) / 5, which simplifies to 2*sqrt(5). So is that the answer? Wait, let me check again.

Alternatively, maybe I made a mistake in the sign when rearranging the equation. Let me check again. Original equation: y = 2x - 2. If I bring all terms to the left side: y - 2x + 2 = 0. Which is equivalent to -2x + y + 2 = 0. So in this case, A is -2, B is 1, C is 2. Then plugging into the formula: |-2*6 + 1*0 + 2| / sqrt((-2)^2 + 1^2). Let me compute that. Numerator: -12 + 0 + 2 = -10, absolute value is 10. Denominator: sqrt(4 + 1) = sqrt(5). So same result, 10 / sqrt(5) = 2*sqrt(5). So same answer. So regardless of which form I use, I get the same result. So that seems good.

But just to be thorough, maybe I can try another method to confirm. For example, using calculus or parametric equations. Let me think. The shortest distance from a point to a line is along the perpendicular. So maybe I can find the equation of the line perpendicular to y = 2x - 2 that passes through (6, 0), find the intersection point with the original line, and then compute the distance between (6, 0) and that intersection point.

Okay, let's try that. The original line has slope 2, so the perpendicular line will have slope -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 me solve for x. Multiply both sides by 2 to eliminate the fraction: 4x - 4 = -x + 6. Then add x to both sides: 5x - 4 = 6. Add 4 to both sides: 5x = 10. Therefore, x = 2. Then plug back into one of the equations to find y. Let's use y = 2x - 2. So y = 2*2 - 2 = 4 - 2 = 2. Therefore, 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] = 2*sqrt(5). So same answer as before. So that confirms it. Therefore, the shortest distance is 2*sqrt(5).

Alternatively, maybe I can use vectors or parametric equations? Let me try parametric equations. Let me parametrize the line y = 2x - 2. Let me let x = t, so y = 2t - 2. So any point on the line can be written as (t, 2t - 2). Then the distance squared from (6, 0) to (t, 2t - 2) is (t - 6)^2 + (2t - 2 - 0)^2. Let me compute that: (t - 6)^2 + (2t - 2)^2. Expand both terms:

First term: (t - 6)^2 = t² - 12t + 36.

Second term: (2t - 2)^2 = 4t² - 8t + 4.

Adding them together: 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 this is a quadratic, the minimum occurs at vertex. The vertex of a quadratic at² + bt + c is at t = -b/(2a). Here, a = 5, b = -20. So t = -(-20)/(2*5) = 20/10 = 2. So t = 2. Then the point on the line is (2, 2*2 - 2) = (2, 2), same 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). Same answer again. So that's consistent.

So all methods lead to the same answer. Therefore, I can be confident that the shortest distance is 2√5.

But just to recap, the formula method was straightforward. The formula is |Ax + By + C| / sqrt(A² + B²). So converting the line equation to standard form, plugging in the point, and simplifying. Then verifying with two other methods: finding the perpendicular line and solving for intersection, and using parametric equations with calculus. All give the same answer. So I think that's solid.

Alternatively, if I didn't remember the formula, I could derive it. Let me try that quickly. Suppose we have a line ax + by + c = 0 and a point (x₀, y₀). The distance from the point to the line is the length of the projection of the vector from any point on the line to (x₀, y₀) onto the normal vector of the line. The normal vector is (a, b). So if we take a point (x₁, y₁) on the line, then the vector from (x₁, y₁) to (x₀, y₀) is (x₀ - x₁, y₀ - y₁). The distance is the absolute value of the dot product of this vector with the unit normal vector.

But since the line equation is ax + by + c = 0, then for any point (x₁, y₁) on the line, ax₁ + by₁ + c = 0. The distance is |ax₀ + by₀ + c| / sqrt(a² + b²). Which is the formula. So that's the derivation. Therefore, the formula is correct.

So in this case, using that formula was the most straightforward. But since I verified with other methods, I can be sure.

Another way to think about it is using coordinates. Let me try to visualize. The point (6, 0) is on the x-axis. The line y = 2x - 2 crosses the y-axis at (0, -2) and the x-axis at (1, 0). So plotting these points, the line goes through (1, 0) and (0, -2). The point (6, 0) is to the right of (1, 0) on the x-axis. The shortest distance should be a perpendicular line from (6, 0) to the given line, which we found to intersect at (2, 2). So the distance between (6, 0) and (2, 2) is sqrt[(6-2)^2 + (0 - 2)^2] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5).

Alternatively, if I didn't know the formula, I could use coordinates and set up equations. Let me try that. Suppose we have a general point (x, y) on the line y = 2x - 2. The distance squared from (6, 0) to (x, y) is (x - 6)^2 + (y - 0)^2. Since y = 2x - 2, substitute that in: (x - 6)^2 + (2x - 2)^2. Which is the same as before. Then take derivative, set to zero, find minimum. Which we did earlier. So same result.

Alternatively, maybe using vectors. The direction vector of the line y = 2x - 2 is (1, 2), since the slope is 2. The vector from (6, 0) to a general point (x, y) on the line is (x - 6, y - 0) = (x - 6, y). For this vector to be perpendicular to the direction vector of the line, their dot product should be zero. So (x - 6, y) • (1, 2) = 0. Which gives (x - 6)*1 + y*2 = 0. But since y = 2x - 2, substitute that in: (x - 6) + 2*(2x - 2) = 0. Let me compute that: x - 6 + 4x - 4 = 0. Combine like terms: 5x - 10 = 0 => 5x = 10 => x = 2. Then y = 2*2 - 2 = 2. So again, the point (2, 2). Then distance is sqrt[(6 - 2)^2 + (0 - 2)^2] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5).

So all methods converge to the same answer. Therefore, I can be confident that the shortest distance is 2√5.

Just to check once more, maybe plugging into the formula. Original line: y = 2x - 2. Let me write it as 2x - y - 2 = 0. So A = 2, B = -1, C = -2. Point (6, 0). Then distance is |2*6 + (-1)*0 - 2| / sqrt(2² + (-1)^2) = |12 - 0 - 2| / sqrt(4 + 1) = |10| / sqrt(5) = 10 / sqrt(5) = 2*sqrt(5). Yep, same result.

Therefore, I think that's solid. The answer is 2√5.

**Final Answer**
The shortest distance 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:

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

---

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

The given line is:

$$
y = 2x - 2
$$

Rewriting it in standard form $Ax + By + C = 0$:

$$
2x - y - 2 = 0
$$

So, we identify:
- $A = 2$
- $B = -1$
- $C = -2$

---

### Step 2: Plug into the distance formula

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

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

---

### Step 3: Simplify the expression

To express the result 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 that the shortest distance from a point to a line is the perpendicular distance. So maybe I can use the formula for the distance from a point to a line? Wait, what's that formula again?

I think it's something like |Ax + By + C| divided by the square root of A squared plus B squared. But let me make sure. The general formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / sqrt(A² + B²). Right, that sounds familiar. So first, I need to write the given line equation in the standard form Ax + By + C = 0.

The given line is y = 2x - 2. Let me rearrange that. Subtract 2x from both sides: -2x + y + 2 = 0. Or maybe I can write it as 2x - y - 2 = 0. Wait, let me check. If I move all terms to the left side: y - 2x + 2 = 0? Wait, no. Let me do 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. So that's the standard form. Alternatively, multiplying both sides by -1 to make the coefficient of x positive: 2x - y - 2 = 0. Either way, both forms are acceptable. Let me pick one. Let me go with 2x - y - 2 = 0. So A is 2, B is -1, and C is -2.

Wait, but in the formula, it's Ax + By + C = 0. So if I have 2x - y - 2 = 0, then A = 2, B = -1, C = -2. So then plugging into the distance formula: |A*x₀ + B*y₀ + C| / sqrt(A² + B²). Let me check that again. So the point is (6, 0). So x₀ is 6, y₀ is 0.

So substituting into the numerator: |2*6 + (-1)*0 + (-2)|. Let me compute that. 2*6 is 12, (-1)*0 is 0, and then -2. So 12 + 0 - 2 = 10. So absolute value of 10 is 10. Then the denominator is sqrt(2² + (-1)^2) which is sqrt(4 + 1) = sqrt(5). Therefore, the distance is 10 / sqrt(5). But they want it in simplest radical form. So rationalizing the denominator: 10 / sqrt(5) is equal to (10*sqrt(5)) / 5, which simplifies to 2*sqrt(5). So is that the answer? Wait, let me check again.

Alternatively, maybe I made a mistake in the sign when rearranging the equation. Let me check again. Original equation: y = 2x - 2. If I bring all terms to the left side: y - 2x + 2 = 0. Which is equivalent to -2x + y + 2 = 0. So in this case, A is -2, B is 1, C is 2. Then plugging into the formula: |-2*6 + 1*0 + 2| / sqrt((-2)^2 + 1^2). Let me compute that. Numerator: -12 + 0 + 2 = -10, absolute value is 10. Denominator: sqrt(4 + 1) = sqrt(5). So same result, 10 / sqrt(5) = 2*sqrt(5). So same answer. So regardless of which form I use, I get the same result. So that seems good.

But just to be thorough, maybe I can try another method to confirm. For example, using calculus or parametric equations. Let me think. The shortest distance from a point to a line is along the perpendicular. So maybe I can find the equation of the line perpendicular to y = 2x - 2 that passes through (6, 0), find the intersection point with the original line, and then compute the distance between (6, 0) and that intersection point.

Okay, let's try that. The original line has slope 2, so the perpendicular line will have slope -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 me solve for x. Multiply both sides by 2 to eliminate the fraction: 4x - 4 = -x + 6. Then add x to both sides: 5x - 4 = 6. Add 4 to both sides: 5x = 10. Therefore, x = 2. Then plug back into one of the equations to find y. Let's use y = 2x - 2. So y = 2*2 - 2 = 4 - 2 = 2. Therefore, 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] = 2*sqrt(5). So same answer as before. So that confirms it. Therefore, the shortest distance is 2*sqrt(5).

Alternatively, maybe I can use vectors or parametric equations? Let me try parametric equations. Let me parametrize the line y = 2x - 2. Let me let x = t, so y = 2t - 2. So any point on the line can be written as (t, 2t - 2). Then the distance squared from (6, 0) to (t, 2t - 2) is (t - 6)^2 + (2t - 2 - 0)^2. Let me compute that: (t - 6)^2 + (2t - 2)^2. Expand both terms:

First term: (t - 6)^2 = t² - 12t + 36.

Second term: (2t - 2)^2 = 4t² - 8t + 4.

Adding them together: 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 this is a quadratic, the minimum occurs at vertex. The vertex of a quadratic at² + bt + c is at t = -b/(2a). Here, a = 5, b = -20. So t = -(-20)/(2*5) = 20/10 = 2. So t = 2. Then the point on the line is (2, 2*2 - 2) = (2, 2), same 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). Same answer again. So that's consistent.

So all methods lead to the same answer. Therefore, I can be confident that the shortest distance is 2√5.

But just to recap, the formula method was straightforward. The formula is |Ax + By + C| / sqrt(A² + B²). So converting the line equation to standard form, plugging in the point, and simplifying. Then verifying with two other methods: finding the perpendicular line and solving for intersection, and using parametric equations with calculus. All give the same answer. So I think that's solid.

Alternatively, if I didn't remember the formula, I could derive it. Let me try that quickly. Suppose we have a line ax + by + c = 0 and a point (x₀, y₀). The distance from the point to the line is the length of the projection of the vector from any point on the line to (x₀, y₀) onto the normal vector of the line. The normal vector is (a, b). So if we take a point (x₁, y₁) on the line, then the vector from (x₁, y₁) to (x₀, y₀) is (x₀ - x₁, y₀ - y₁). The distance is the absolute value of the dot product of this vector with the unit normal vector.

But since the line equation is ax + by + c = 0, then for any point (x₁, y₁) on the line, ax₁ + by₁ + c = 0. The distance is |ax₀ + by₀ + c| / sqrt(a² + b²). Which is the formula. So that's the derivation. Therefore, the formula is correct.

So in this case, using that formula was the most straightforward. But since I verified with other methods, I can be sure.

Another way to think about it is using coordinates. Let me try to visualize. The point (6, 0) is on the x-axis. The line y = 2x - 2 crosses the y-axis at (0, -2) and the x-axis at (1, 0). So plotting these points, the line goes through (1, 0) and (0, -2). The point (6, 0) is to the right of (1, 0) on the x-axis. The shortest distance should be a perpendicular line from (6, 0) to the given line, which we found to intersect at (2, 2). So the distance between (6, 0) and (2, 2) is sqrt[(6-2)^2 + (0 - 2)^2] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5).

Alternatively, if I didn't know the formula, I could use coordinates and set up equations. Let me try that. Suppose we have a general point (x, y) on the line y = 2x - 2. The distance squared from (6, 0) to (x, y) is (x - 6)^2 + (y - 0)^2. Since y = 2x - 2, substitute that in: (x - 6)^2 + (2x - 2)^2. Which is the same as before. Then take derivative, set to zero, find minimum. Which we did earlier. So same result.

Alternatively, maybe using vectors. The direction vector of the line y = 2x - 2 is (1, 2), since the slope is 2. The vector from (6, 0) to a general point (x, y) on the line is (x - 6, y - 0) = (x - 6, y). For this vector to be perpendicular to the direction vector of the line, their dot product should be zero. So (x - 6, y) • (1, 2) = 0. Which gives (x - 6)*1 + y*2 = 0. But since y = 2x - 2, substitute that in: (x - 6) + 2*(2x - 2) = 0. Let me compute that: x - 6 + 4x - 4 = 0. Combine like terms: 5x - 10 = 0 => 5x = 10 => x = 2. Then y = 2*2 - 2 = 2. So again, the point (2, 2). Then distance is sqrt[(6 - 2)^2 + (0 - 2)^2] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5).

So all methods converge to the same answer. Therefore, I can be confident that the shortest distance is 2√5.

Just to check once more, maybe plugging into the formula. Original line: y = 2x - 2. Let me write it as 2x - y - 2 = 0. So A = 2, B = -1, C = -2. Point (6, 0). Then distance is |2*6 + (-1)*0 - 2| / sqrt(2² + (-1)^2) = |12 - 0 - 2| / sqrt(4 + 1) = |10| / sqrt(5) = 10 / sqrt(5) = 2*sqrt(5). Yep, same result.

Therefore, I think that's solid. The answer is 2√5.

**Final Answer**
The shortest distance 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:

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

---

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

The given line is:

$$
y = 2x - 2
$$

Rewriting it in standard form $Ax + By + C = 0$:

$$
2x - y - 2 = 0
$$

So, we identify:
- $A = 2$
- $B = -1$
- $C = -2$

---

### Step 2: Plug into the distance formula

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

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

---

### Step 3: Simplify the expression

To express the result 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: $$\sqrt{(6-2)^2+(0-2)^2}=\sqrt{16+4}=\boxed{2\sqrt{5}}$$