Question

1. triangle pqr with vertices p(-4, 4), q(0, 2), and r(-1, -3) is reflected over a line to form triangle pqr with vertices p(6, 4), q(2, 2) and r(3, -3). what is the equation of the line over which triangle pqr was reflected?

Explanation:

Step1: Find mid - points

Mid - point of $PP'$: $(\frac{-4 + 6}{2},\frac{4+4}{2})=(1,4)$. Mid - point of $QQ'$: $(\frac{0 + 2}{2},\frac{2 + 2}{2})=(1,2)$. Mid - point of $RR'$: $(\frac{-1+3}{2},\frac{-3-3}{2})=(1,-3)$.

Step2: Determine the line

The line of reflection is the vertical line passing through the common $x$ - coordinate of mid - points. So the equation is $x = 1$.

Answer:

$x = 1$