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: Recall mid - point formula

The mid - point between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. The line of reflection is the perpendicular bisector of the line segment joining a point and its image.

Step2: Find mid - point of $PP'$

For $P(-4,4)$ and $P'(6,4)$, the mid - point $M_1$ is $(\frac{-4 + 6}{2},\frac{4+4}{2})=(1,4)$.

Step3: Find mid - point of $QQ'$

For $Q(0,2)$ and $Q'(2,2)$, the mid - point $M_2$ is $(\frac{0 + 2}{2},\frac{2+2}{2})=(1,2)$.

Step4: Determine the line of reflection

Since the $y$ - coordinates of the mid - points of $PP'$ and $QQ'$ are different but the $x$ - coordinates are the same ($x = 1$), and the transformation is a reflection, the equation of the line of reflection is $x=1$.

Answer:

$x = 1$