Question

a 90° rotation around the origin followed by a reflection across the line x = 3 maps △jkl, with coordinates j(-2, 2), k(-2, 6), and l(-8, 2), to △pqr. find the area of △pqr. square units

Explanation:

Step1: Find the area of △JKL

First, we can use the formula for the area of a triangle with coordinates. The coordinates of J(-2, 2), K(-2, 6), and L(-8, 2). Notice that J and K have the same x - coordinate, so the length of JK is the difference in y - coordinates. \(JK=\vert6 - 2\vert=4\). J and L have the same y - coordinate, so the length of JL (the base or height, depending on how we look at it) is the difference in x - coordinates. \(JL=\vert-2-(-8)\vert = 6\). Since △JKL is a right triangle (because JK is vertical and JL is horizontal), the area of a right triangle is \(A=\frac{1}{2}\times base\times height\). Here, base = JL = 6 and height = JK = 4. So the area of △JKL is \(\frac{1}{2}\times6\times4 = 12\).

Step2: Determine the area of △PQR

Rotations and reflections are rigid transformations. Rigid transformations (rotations, reflections, translations) do not change the shape or size of a figure, which means they preserve the area. So the area of △PQR is the same as the area of △JKL.

Answer:

12