Question

suppose m is the line with the equation y = 3. given a(7,1), b(2,9), and c(3, - 5), what are the coordinates of the vertices of △abc for $(t_{(-4,0)} circ r_m)(\triangle abc) = \triangle abc$? choose the correct answer below. a. $a(3,5), b(-2,-3), c(-1,11)$ b. $a(3,4), b(-2,12), c(-1,-2)$ c. $a(11,4), b(6,12), c(7,-2)$ d. $a(11,5), b(-6,-3), c(7,11)$

Explanation:

Step1: Understand the Transformations

First, we have a reflection over the line \( y = 3 \) (denoted as \( r_m \)) followed by a translation \( T_{(-4,0)} \) (which means we subtract 4 from the \( x \)-coordinate and leave the \( y \)-coordinate unchanged).

Step2: Reflect a Point over \( y = 3 \)

The formula for reflecting a point \( (x,y) \) over the horizontal line \( y = k \) is \( (x, 2k - y) \). Here, \( k = 3 \), so the reflection of a point \( (x,y) \) over \( y = 3 \) is \( (x, 6 - y) \).

Reflect Point A(7,1):

Using the reflection formula: \( x = 7 \), \( y = 1 \). So the reflected point \( A_1 \) (after reflection) is \( (7, 6 - 1) = (7, 5) \).

Reflect Point B(2,9):

Reflected point \( B_1 \) is \( (2, 6 - 9) = (2, -3) \).

Reflect Point C(3,-5):

Reflected point \( C_1 \) is \( (3, 6 - (-5)) = (3, 11) \).

Step3: Translate the Reflected Points by \( T_{(-4,0)} \)

The translation \( T_{(-4,0)} \) means we apply the transformation \( (x - 4, y + 0) \) to the reflected points.

Translate \( A_1(7,5) \):

New \( x \)-coordinate: \( 7 - 4 = 3 \), \( y \)-coordinate: \( 5 \). So \( A' = (3, 5) \)? Wait, no, wait. Wait, maybe I made a mistake. Wait, the order of transformations: \( (T_{(-4,0)} \circ r_m)(\triangle ABC) \) means first reflect over \( m \), then translate by \( T_{(-4,0)} \). Wait, but let's check the answer options. Wait, option B has \( A'(3,4) \), so maybe my reflection formula is wrong? Wait, no, wait, maybe the line is \( y = 3 \), let's recheck.

Wait, maybe the reflection formula is \( (x, 2k - y) \). For \( y = 3 \), \( 2k = 6 \), so \( 6 - y \). Let's check point A(7,1): \( 6 - 1 = 5 \), so reflected point is (7,5). Then translate by \( (-4,0) \): \( (7 - 4, 5) = (3,5) \). But option B has \( A'(3,4) \). Wait, maybe I mixed up the order of transformations? Wait, the notation \( (T_{(-4,0)} \circ r_m) \) means we first do \( r_m \), then \( T_{(-4,0)} \). Wait, maybe the problem is \( (r_m \circ T_{(-4,0)}) \)? No, the notation is \( (T_{(-4,0)} \circ r_m) \), which is function composition: first apply \( r_m \), then apply \( T_{(-4,0)} \).

Wait, let's check option B: \( A'(3,4) \), \( B'(-2,12) \), \( C'(-1,-2) \). Let's try another approach. Maybe the reflection is over \( y = 3 \), but maybe I miscalculated. Wait, let's take point A(7,1). The distance from \( y = 1 \) to \( y = 3 \) is \( 3 - 1 = 2 \), so the reflection should be 2 units above \( y = 3 \), so \( y = 3 + 2 = 5 \). So reflected point is (7,5). Then translate by (-4,0): \( 7 - 4 = 3 \), \( y = 5 \), so (3,5). But option A has \( A'(3,5) \), \( B'(-2,-3) \), \( C'(-1,11) \). Wait, let's check point B(2,9). Distance from \( y = 9 \) to \( y = 3 \) is \( 9 - 3 = 6 \), so reflection is 6 units below \( y = 3 \), so \( y = 3 - 6 = -3 \). So reflected point is (2, -3). Then translate by (-4,0): \( 2 - 4 = -2 \), \( y = -3 \), so \( B' = (-2, -3) \). Point C(3,-5): distance from \( y = -5 \) to \( y = 3 \) is \( 3 - (-5) = 8 \), so reflection is 8 units above \( y = 3 \), so \( y = 3 + 8 = 11 \). Reflected point is (3,11). Translate by (-4,0): \( 3 - 4 = -1 \), \( y = 11 \), so \( C' = (-1, 11) \). So the coordinates are \( A'(3,5) \), \( B'(-2,-3) \), \( C'(-1,11) \), which is option A. Wait, but the original selected option was B. Wait, maybe I messed up the order of transformations. Wait, the notation \( (T_{(-4,0)} \circ r_m) \) is "first \( r_m \), then \( T_{(-4,0)} \)". But maybe the problem is \( (r_m \circ T_{(-4,0)}) \)? Let's check that.

If we first translate by \( T_{(-4,0)} \), then reflect over \( y = 3 \). Let's try that.

Translate A(7,1) by \( T_{(-4,0)} \): \( (7 - 4, 1) = (3,1) \). Then reflect over \( y = 3 \): \( (3, 6 - 1) = (3,5) \). No, same as before.

Wait, option B: \( A'(3,4) \). Let's see, maybe the reflection is over \( y = 2 \)? No, the line is \( y = 3 \). Wait, maybe my reflection formula is incorrect. The formula for reflecting over \( y = k \) is \( (x, 2k - y) \). Let's verify with a point. For example, reflecting (x, k) over \( y = k \) should give (x, k), since \( 2k - k = k \). Correct. Reflecting (x, k + d) over \( y = k \) should give (x, k - d), since \( 2k - (k + d) = k - d \). Correct. So for (7,1), \( k = 3 \), \( d = 2 \) (since \( 3 - 1 = 2 \)), so reflection is (7, 3 - 2) = (7,1)? No, wait, no: \( 2k - y = 6 - 1 = 5 \), which is \( 3 + 2 \), so it's 2 units above \( y = 3 \). So (7,5) is correct. Then translating by (-4,0) gives (3,5). Which is option A's \( A' \). Then \( B(2,9) \): \( 2k - y = 6 - 9 = -3 \), so (2,-3). Translate by (-4,0): (2 - 4, -3) = (-2, -3), which is option A's \( B' \). Then \( C(3,-5) \): \( 2k - y = 6 - (-5) = 11 \), so (3,11). Translate by (-4,0): (3 - 4, 11) = (-1,11), which is option A's \( C' \). So the correct answer should be option A. Wait, but the original selected option was B. Maybe there was a mistake in the problem or in my understanding. Wait, let's check the answer options again. Option A: \( A'(3,5), B'(-2,-3), C'(-1,11) \). Let's verify the transformations again.

First, reflection over \( y = 3 \):

- A(7,1): distance to \( y = 3 \) is \( 3 - 1 = 2 \), so reflection is \( 3 + 2 = 5 \), so (7,5). Then translate by (-4,0): \( 7 - 4 = 3 \), \( y = 5 \), so (3,5). Correct.

- B(2,9): distance to \( y = 3 \) is \( 9 - 3 = 6 \), so reflection is \( 3 - 6 = -3 \), so (2,-3). Translate by (-4,0): \( 2 - 4 = -2 \), \( y = -3 \), so (-2,-3). Correct.

- C(3,-5): distance to \( y = 3 \) is \( 3 - (-5) = 8 \), so reflection is \( 3 + 8 = 11 \), so (3,11). Translate by (-4,0): \( 3 - 4 = -1 \), \( y = 11 \), so (-1,11). Correct.

So the correct answer is option A. Wait, but the original selected option was B. Maybe the problem was written with the translation first? Let's check option B: \( A'(3,4) \). Let's see, if we first translate A(7,1) by (-4,0) to get (3,1), then reflect over \( y = 3 \): \( 6 - 1 = 5 \), no, that's (3,5). Not (3,4). So maybe the line is \( y = 2 \)? If \( k = 2 \), then reflection of (7,1) is (7, 4 - 1) = (7,3), then translate by (-4,0): (3,3). No. Alternatively, maybe the translation is \( T_{(4,0)} \)? No, the problem says \( T_{(-4,0)} \). So based on the calculations, the correct answer is option A.

Answer:

A. \( A'(3,5), B'(-2,-3), C'(-1,11) \)