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∆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?
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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)
to reflect the whole triangle about a line, reflect the vertices.
idk how to do it this what im doing is all computerized
so let's do this step-wise... A'(-1,1) reflected along the line y = ix.|dw:1396098595365:dw|
yes
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All right. So one point is on the line itself, so it stays there.
yes
another thing you should note that each point is equidistant from the translated point.
yeah i see that
do you know what the distance of a point from a line is given by?
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