Question

reflecting other lines reflect triangle jkl across the line x = 1. a. draw line segments connecting the vertices of the pre - image and image. b. what do you notice about these line segments and the line of reflection?

Explanation:

Step1: Recall reflection rule

For a point $(x,y)$ reflected across the line $x = a$, the new - point is $(2a - x,y)$. Here $a = 1$.

Step2: Identify vertices of $\triangle{JKL}$

Let's assume the coordinates of $J=(4,5)$, $K=(1,7)$, $L=(2,3)$.

Step3: Calculate reflected vertices

For point $J=(4,5)$: $x'=2\times1 - 4=- 2$, $y' = 5$, so $J'=(-2,5)$.
For point $K=(1,7)$: $x'=2\times1 - 1 = 1$, $y' = 7$, so $K'=(1,7)$.
For point $L=(2,3)$: $x'=2\times1 - 2 = 0$, $y' = 3$, so $L'=(0,3)$.

Step4: Answer part a

Draw the line segments connecting the vertices of $\triangle{JKL}$ (with vertices $J(4,5)$, $K(1,7)$, $L(2,3)$) and the line segments connecting the vertices of $\triangle{J'K'L'}$ (with vertices $J'(-2,5)$, $K'(1,7)$, $L'(0,3)$).

Step5: Answer part b

The line of reflection $x = 1$ is the perpendicular bisector of each line segment connecting a vertex of the pre - image and its corresponding vertex of the image. Also, the distance from each vertex of the pre - image to the line of reflection is equal to the distance from its corresponding vertex of the image to the line of reflection.

Answer:

a. Plot $\triangle{JKL}$ with $J(4,5)$, $K(1,7)$, $L(2,3)$ and $\triangle{J'K'L'}$ with $J'(-2,5)$, $K'(1,7)$, $L'(0,3)$ and draw the line segments connecting the vertices of each triangle.
b. The line of reflection $x = 1$ is the perpendicular bisector of the line segments connecting corresponding vertices of the pre - image and image, and the distances from corresponding vertices to the line of reflection are equal.