∆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?

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?

anonymous · Answer

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)

parthkohli · Answer

to reflect the whole triangle about a line, reflect the vertices.

anonymous · Answer

idk how to do it this what im doing is all computerized

parthkohli · Answer

so let's do this step-wise...

A'(-1,1) reflected along the line y = ix.|dw:1396098595365:dw|

anonymous · Answer

yes

parthkohli · Answer

All right.  So one point is on the line itself, so it stays there.

anonymous · Answer

yes

parthkohli · Answer

another thing you should note that each point is equidistant from the translated point.

anonymous · Answer

yeah i see that

parthkohli · Answer

do you know what the distance of a point from a line is given by?