Question

question 12
1 pts
which set of coordinates provide the vertices for a 90° rotation of △abc about point b?
graph of coordinate plane with triangle abc
options:
a(-1, 4), b(-1, 2), c(2, 2)
a(1, 2), b(3, 2), c(3, 5)
a(3, 2), b(1, 2), c(1, 5)
a(-3, 3), b(-1, 2), c(0, 5)

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of \( \triangle ABC \). From the graph: \( A(-3, 2) \), \( B(-1, 2) \), \( C(-1, 5) \).

Step2: Recall 90° rotation about a point

To rotate a point \( (x, y) \) 90° counterclockwise about a point \( (a, b) \), the transformation is \( (x - a, y - b) \to (- (y - b), x - a) \), then add \( (a, b) \) back. For 90° clockwise, it's \( (x - a, y - b) \to (y - b, - (x - a)) \), then add \( (a, b) \). Let's check the rotation about \( B(-1, 2) \).

For point \( A(-3, 2) \): Vector from \( B \) to \( A \) is \( (-3 - (-1), 2 - 2) = (-2, 0) \). Rotating 90° counterclockwise: \( (0, -2) \) (since \( (x, y) \to (-y, x) \)). Then add \( B \)'s coordinates: \( (-1 + 0, 2 + (-2)) = (-1, 0) \)? Wait, maybe clockwise. Wait, let's check the options. Let's look at the vectors.

Original \( A(-3,2) \), \( B(-1,2) \), \( C(-1,5) \). The vector \( \overrightarrow{BA} = (-3 - (-1), 2 - 2) = (-2, 0) \), \( \overrightarrow{BC} = (-1 - (-1), 5 - 2) = (0, 3) \).

Rotating 90° clockwise: \( \overrightarrow{BA} \) becomes \( (0, 2) \) (since \( (x, y) \to (y, -x) \): \( (-2, 0) \to (0, 2) \)), so new \( A \) is \( B + (0, 2) = (-1, 2 + 2) = (-1, 4) \). \( \overrightarrow{BC} \) rotating 90° clockwise: \( (0, 3) \to (3, 0) \) (since \( (x, y) \to (y, -x) \): \( (0, 3) \to (3, 0) \)), so new \( C \) is \( B + (3, 0) = (-1 + 3, 2 + 0) = (2, 2) \). So new \( A(-1,4) \), \( B(-1,2) \), \( C(2,2) \), which is the first option.

Wait, let's verify the first option: \( A(-1,4) \), \( B(-1,2) \), \( C(2,2) \). The vector from \( B \) to \( A \) is \( (0, 2) \), from \( B \) to \( C \) is \( (3, 0) \). The original vectors were \( (-2, 0) \) and \( (0, 3) \). Rotating 90° clockwise: \( (-2, 0) \) becomes \( (0, 2) \) (since \( (x,y) \to (y, -x) \): \( (-2,0) \to (0, 2) \)), and \( (0,3) \) becomes \( (3, 0) \) ( \( (0,3) \to (3, 0) \) ). So adding to \( B(-1,2) \), we get \( A(-1 + 0, 2 + 2) = (-1,4) \), \( C(-1 + 3, 2 + 0) = (2,2) \). So the first option matches.

Step3: Check other options

Second option: \( A(1,2) \), \( B(3,2) \), \( C(3,5) \). Vector \( \overrightarrow{BA} = (1 - 3, 2 - 2) = (-2, 0) \), \( \overrightarrow{BC} = (3 - 3, 5 - 2) = (0, 3) \). Rotating about \( B(3,2) \)? No, original \( B \) is \( (-1,2) \), so this is not.

Third option: \( A(3,2) \), \( B(1,2) \), \( C(1,5) \). Vector \( \overrightarrow{BA} = (3 - 1, 2 - 2) = (2, 0) \), \( \overrightarrow{BC} = (1 - 1, 5 - 2) = (0, 3) \). Rotating about \( B(1,2) \), but original \( B \) is \( (-1,2) \), so no.

Fourth option: \( A(-3,3) \), \( B(-1,2) \), \( C(0,5) \). Vectors don't match the rotation.

So the correct option is the first one.

Answer:

A. \( A(-1, 4) \), \( B(-1, 2) \), \( C(2, 2) \)