Question

reflecting a triangle consider reflections of △jkl. what line of reflection maps point k to point k at (-5, 2)? what line of reflection maps point l to point l at (-2, 3)?

Explanation:

Step1: Find mid - point for K and K'

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $K(5,2)$ and $K'(-5,2)$, the mid - point is $(\frac{5+( - 5)}{2},\frac{2 + 2}{2})=(0,2)$. The line of reflection is the vertical line passing through the mid - point, so the line is $x = 0$.

Step2: Find mid - point for L and L'

For $L(-3,2)$ and $L'(-2,3)$, the mid - point is $(\frac{-3+( - 2)}{2},\frac{2 + 3}{2})=(-\frac{5}{2},\frac{5}{2})$. The slope of the line passing through $L$ and $L'$ is $m=\frac{3 - 2}{-2-( - 3)} = 1$. The slope of the perpendicular bisector (line of reflection) is $-1$. Using the point - slope form $y - y_0=m(x - x_0)$ with the mid - point $(-\frac{5}{2},\frac{5}{2})$ and $m=-1$, we have $y-\frac{5}{2}=-1(x+\frac{5}{2})$, which simplifies to $y=-x$.

Answer:

The line of reflection that maps point K to point K' is $x = 0$.
The line of reflection that maps point L to point L' is $y=-x$.