<Question>
A reflection takes $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 4 \\ 3 \end{pmatrix}.$  Which vector does the reflection take $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to?
</Question>

<think>
<Plan_1>
Okay, let's try to figure out this reflection problem. So, we know that the reflection takes the vector (5, 0) to (4, 3). We need to find out which vector the reflection takes (-2, 3) to. Hmm, reflections... right, reflections are linear transformations that flip points over a line in the plane. The line of reflection is the invariant line, meaning any point on that line stays the same after reflection. So, first, I need to find the equation of the line of reflection.

But how do we find the line of reflection given that (5, 0) maps to (4, 3)? Let me think. Since reflection is its own inverse, the midpoint between the original point and its image should lie on the line of reflection. That makes sense because the line of reflection is the perpendicular bisector of the segment joining a point and its image. So, let's compute the midpoint between (5, 0) and (4, 3).

<prompt_1.1> Compute the midpoint between the points (5, 0) and (4, 3) to find the line of reflection. </prompt_1.1>
-----

Alternatively, maybe there's another way to approach this. Let's see. Let's denote the reflection matrix as R. Then, R*(5,0) = (4,3). We need to find R*(-2,3). But to do that, we need to find the matrix R. But since we only have one vector transformed, maybe we can determine R?

<prompt_1.2> Consider an alternative method to find the reflection matrix R. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
The midpoint formula is ((x1 + x2)/2, (y1 + y2)/2). Plugging in the values: ((5 + 4)/2, (0 + 3)/2) = (9/2, 3/2). So the midpoint is (4.5, 1.5). That point lies on the line of reflection.

Also, the line of reflection is perpendicular to the line joining the original point and its image. Let's find the slope of the line joining (5, 0) and (4, 3). The slope is (3 - 0)/(4 - 5) = 3/(-1) = -3. Therefore, the slope of the line of reflection, being perpendicular, is the negative reciprocal, which is 1/3.
</execution_1.1>

<execution_1.2>
Wait, reflection matrices are determined by their line of reflection. So, if we can find the line of reflection, then we can find the matrix. Alternatively, maybe we can use the formula for reflection over a line. The general formula for reflection over a line making an angle θ with the x-axis is:

R = [cos2θ   sin2θ]
[sin2θ  -cos2θ]

Wait, is that right? Let me recall. The standard reflection matrix over a line at angle θ is:

[cos2θ   sin2θ]
[sin2θ  -cos2θ]

Yes, that seems correct. So, if we can find θ, then we can compute the reflection matrix.

Alternatively, another approach is to use the midpoint and the fact that the line of reflection is the perpendicular bisector. Let's try this geometric approach first.

So, the midpoint between (5, 0) and (4, 3) is ((5+4)/2, (0+3)/2) = (9/2, 3/2). This midpoint lies on the line of reflection. Also, the line of reflection is perpendicular to the line joining (5, 0) and (4, 3). Let's compute the slope of the line joining these two points.

The original point is (5, 0) and the image is (4, 3). The slope of the line connecting them is (3 - 0)/(4 - 5) = 3/(-1) = -3. Therefore, the line of reflection, being perpendicular, has a slope of 1/3 (since the negative reciprocal of -3 is 1/3).
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So, we have a point (4.5, 1.5) on the line and a slope of 1/3. Therefore, the equation of the line of reflection is y - 1.5 = (1/3)(x - 4.5). Let me simplify that.

<prompt_2.1> Determine the equation of the line of reflection using the midpoint and slope. </prompt_2.1>
-----

Based on execution_1.2:
So, the line of reflection has slope 1/3 and passes through the midpoint (4.5, 1.5). Let's confirm that. Wait, actually, we found the slope from the perpendicularity, and we know a point on the line (the midpoint). Therefore, the equation is y = (1/3)x + b. Plugging in x = 4.5, y = 1.5:

<prompt_2.2> Confirm the equation of the line of reflection using the midpoint and slope. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Multiply both sides by 3 to eliminate the fraction: 3y - 4.5 = x - 4.5. Then, subtract x from both sides: -x + 3y - 4.5 = -4.5. Adding 4.5 to both sides gives -x + 3y = 0, or equivalently, x - 3y = 0. Wait, is that correct? Let me check again.

Starting from y - 1.5 = (1/3)(x - 4.5). Let's distribute the 1/3: y - 1.5 = (1/3)x - 1.5. Then, adding 1.5 to both sides: y = (1/3)x. So, the equation simplifies to y = (1/3)x. Wait, that's interesting. So the line of reflection is y = (1/3)x?
</execution_2.1>

<execution_2.2>
1.5 = (1/3)(4.5) + b
1.5 = 1.5 + b
So, b = 0. Therefore, the equation of the line of reflection is y = (1/3)x.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But wait, if the midpoint is (4.5, 1.5), then plugging x = 4.5 into y = (1/3)x gives y = 1.5.
-----

Based on execution_1.2, execution_2.2:
Now, to find the reflection of a point over the line y = (1/3)x. Let's recall the formula for reflection over a line ax + by + c = 0. The reflection of a point (x, y) is given by:

<prompt_3.1> Use the reflection formula over the line y = (1/3)x to find the reflection of the point (-2, 3). </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
(x - 2a(ax + by + c)/(a² + b²), y - 2b(ax + by + c)/(a² + b²))

But maybe it's easier to use the formula for reflection over y = mx + c. Since our line is y = (1/3)x, which can be written as (1/3)x - y = 0. So, m = 1/3, c = 0.

The formula for reflection over y = mx is:

x' = ( (1 - m²)x + 2m y ) / (1 + m²)
y' = ( 2m x + (m² - 1) y ) / (1 + m²)

Let me verify this formula. Yes, I think that's correct. Let me check with a simple example. Suppose m = 0, reflection over the x-axis. Then,

x' = (1 - 0)x + 0*y / 1 = x
y' = 0*x + (-1)y / 1 = -y
Which is correct. If m is infinity (vertical line), but in our case m = 1/3.

So, using this formula, let's plug in m = 1/3. Then,

x' = (1 - (1/3)^2)x + 2*(1/3)y / (1 + (1/3)^2)
y' = 2*(1/3)x + ((1/3)^2 - 1)y / (1 + (1/3)^2)

Compute denominators first: 1 + (1/3)^2 = 1 + 1/9 = 10/9.

So, denominators are 10/9, so we can multiply numerator and denominator accordingly.

Compute x':

Numerator for x' : (1 - 1/9)x + (2/3)y = (8/9)x + (2/3)y
Divide by 10/9: (8/9)x + (2/3)y multiplied by 9/10 = (8x + 6y)/10 = (4x + 3y)/5

Similarly, y':

Numerator for y' : (2/3)x + (1/9 - 1)y = (2/3)x + (-8/9)y
Divide by 10/9: (2/3)x + (-8/9)y multiplied by 9/10 = (6x - 8y)/10 = (3x - 4y)/5

Therefore, the reflection matrix over y = (1/3)x is:

R = [4/5   3/5]
[3/5  -4/5]

Wait, let me confirm the signs. For y', the numerator was (2/3)x + (-8/9)y. Then multiplying by 9/10:


-----

So, let's confirm this. Let's apply this reflection matrix to the vector (5, 0):
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
First component: (4*5 + 3*0)/5 = 20/5 = 4

Second component: (3*5 + (-4)*0)/5 = 15/5 = 3

Wait, but the image is supposed to be (4, 3). Wait, that's exactly the given image. So, applying the reflection matrix to (5, 0) gives (4, 3), which matches the problem statement. So, this seems correct.

Therefore, the reflection matrix is indeed R = [4/5  3/5; 3/5  -4/5].

Based on execution_3.1:
Now, we need to apply this reflection matrix to the vector (-2, 3). Let's compute each component.

<prompt_4.1> Apply the reflection matrix to the vector (-2, 3) to find the reflected vector. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
First component: (4*(-2) + 3*3)/5 = (-8 + 9)/5 = 1/5 = 0.2

Second component: (3*(-2) + (-4)*3)/5 = (-6 -12)/5 = (-18)/5 = -3.6

So, the reflected vector is (1/5, -18/5). Let's write that as fractions: (1/5, -18/5). Hmm, that seems correct, but let's verify.

Alternatively, maybe there's another way to compute this. Let's use the formula for reflection over a line. Let me recall the formula again. For a line y = mx, the reflection of a point (x, y) is:

( (x + m y)/(1 + m²), (m x + y)/(1 + m²) )

Wait, no, that's not quite. Wait, actually, the formula I derived earlier was:

x' = ( (1 - m²)x + 2m y ) / (1 + m²)
y' = ( 2m x + (m² - 1) y ) / (1 + m²)

So, plugging m = 1/3, x = -2, y = 3:

x' = ( (1 - (1/3)^2)*(-2) + 2*(1/3)*3 ) / (1 + (1/3)^2 )
y' = ( 2*(1/3)*(-2) + ( (1/3)^2 - 1 )*3 ) / (1 + (1/3)^2 )

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

Compute x':

Numerator: (1 - 1/9)*(-2) + (2/3)*3 = (8/9)*(-2) + 2 = (-16/9) + 2 = (-16/9) + (18/9) = 2/9

Then, x' = (2/9) / (10/9) = 2/10 = 1/5

Compute y':

Numerator: (2/3)*(-2) + (1/9 - 1)*3 = (-4/3) + (-8/9)*3 = (-4/3) + (-24/9) = (-4/3) - 8/3 = (-12/3) = -4

Then, y' = (-4) / (10/9) = (-4)*(9/10) = -36/10 = -18/5

So, the reflection is (1/5, -18/5), which matches the previous result. Therefore, the vector (-2, 3) is reflected to (1/5, -18/5).
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But let's write that as fractions: (1/5, -18/5). Hmm, seems correct. Let me check once more.

Alternatively, let's use geometric reasoning.

<prompt_5.1> Verify the reflection result using geometric reasoning. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Since the line of reflection is y = (1/3)x, and we are reflecting the point (-2, 3). Let's see if this point and its reflection are equidistant from the line and if the line is the perpendicular bisector.

Alternatively, maybe we can parametrize the reflection. Let me consider the formula again.

But given that both methods gave the same answer, and the verification with the midpoint and slope seems correct, I think this is the right answer.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Let me confirm with another approach. Let's compute the reflection using vector projection.

<prompt_6.1> Confirm the reflection result using vector projection and verify the result. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
The formula for reflection over a line can be expressed as:

Reflection = 2 * projection - original vector

So, if we can find the projection of the vector (-2, 3) onto the line y = (1/3)x, then the reflection is 2*projection - original vector.

First, let's find the projection. The line y = (1/3)x has direction vector (3, 1) because the slope is 1/3. So, a direction vector is (3, 1). To project the vector v = (-2, 3) onto the line, we can use the projection formula:

proj_L(v) = ( (v ⋅ ī) / ||ī||² ) * ī

where ī is a direction vector of the line. Let's take ī = (3, 1).

Compute v ⋅ ī = (-2)(3) + (3)(1) = -6 + 3 = -3

||ī||² = 3² + 1² = 9 + 1 = 10

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

Then, the reflection vector is 2*proj_L(v) - v = 2*(-9/10, -3/10) - (-2, 3) = ( -18/10, -6/10 ) + (2, -3 ) = ( -18/10 + 20/10, -6/10 - 30/10 ) = (2/10, -36/10 ) = (1/5, -18/5 )

Same result! So, this confirms that the reflection of (-2, 3) over the line y = (1/3)x is indeed (1/5, -18/5).
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Therefore, the answer should be \boxed{\begin{pmatrix} \dfrac{1}{5} \\ -\dfrac{18}{5} \end{pmatrix}}.

But let me just cross-verify once more with the midpoint.

<prompt_7.1> Perform final cross-verification using the midpoint and confirm the answer. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Suppose we reflect (-2, 3). The midpoint between (-2, 3) and its reflection (1/5, -18/5) should lie on the line y = (1/3)x.

Compute the midpoint: ( (-2 + 1/5)/2, (3 + (-18/5))/2 )

First component: (-10/5 + 1/5)/2 = (-9/5)/2 = -9/10

Second component: (15/5 - 18/5)/2 = (-3/5)/2 = -3/10

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

Plug x = -9/10 into the line equation: y = (1/3)(-9/10) = -3/10. Which matches the y-coordinate of the midpoint. Perfect, that confirms that the reflection is correct.

Another check: The line connecting (-2, 3) and (1/5, -18/5) should be perpendicular to the line y = (1/3)x. Let's compute the slope of the connecting line.

Slope = ( -18/5 - 3 ) / (1/5 - (-2)) = ( -18/5 - 15/5 ) / (1/5 + 10/5 ) = ( -33/5 ) / (11/5 ) = -33/5 * 5/11 = -33/11 = -3

Which is indeed the negative reciprocal of 1/3, confirming perpendicularity. All checks out!

Therefore, after thorough analysis and multiple methods confirming the result, the reflection of (-2, 3) is (1/5, -18/5).
</execution_7.1>
</Execution_7>

<Final_answer>
Therefore, the vector which the reflection takes $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ to is $\boxed{\begin{pmatrix} \frac{1}{5} \\ -\frac{18}{5} \end{pmatrix}}$.
</Final_answer>
</think>