Question

∆ABC is reflected about the line y = -x to give ∆A'B'C' with vertices A'(-1, 1), B'(-2, -1), C(-1, 0). What are the vertices of ∆ABC? A(1, -1), B(-1, -2), C(0, -1) A(-1, 1), B(1, 2), C(0, 1) A(-1, -1), B(-2, -1), C(-1, 0) A(1, 1), B(2, -1), C(1, 0) A(1, 2), B(-1, 1), C(0, 1) NextReset

Answer 1

The changes in coordinate that undergo diagonal line reflection would be switching between the x and y coordinate. In y=-x diagonal line, the change will also apply minus. You are given the reflected vertice and asked the real triangle coordinate. Then you just need to apply the transformation change. The coordinate would be:

Ax= A'y * -1= 1*-1 = -1

Ay= A'x * -1= -1 * -1 = 1

A= (-1, 1)

Bx= B'y * -1= -1*-1 = 1

By= B'x * -1= -2 * -1 = 2

B= (1, 2)

Cx= C'y * -1= 0*-1 = 0

Cy= C'x * -1= -1 * -1 = 1

C= (0, 1)

Explanation:

Answer 2

Answer: A= (-1, 1) B= (1, 2) C= (0, 1)

The changes in coordinate that undergo diagonal line reflection would be switching between the x and y coordinate. In y=-x diagonal line, the change will also apply minus. You are given the reflected vertice and asked the real triangle coordinate. Then you just need to apply the transformation change. The coordinate would be:

Ax= A'y * -1= 1*-1 = -1
Ay= A'x * -1= -1 * -1 = 1
A= (-1, 1)

Bx= B'y * -1= -1*-1 = 1
By= B'x * -1= -2 * -1 = 2
B= (1, 2)

Cx= C'y * -1= 0*-1 = 0
Cy= C'x * -1= -1 * -1 = 1
C= (0, 1)