Question

c) a reflection in the line y = -1
point a is 4 units above y = -1, so a is 4 units below y = -1 at (1, -5).
similarly, b is 3 units below y = -1 at (5, -4), and c is 2 units below y = -1 at (2, -3).

visit www.bigideasmathvideos.com to watch the flipped video instruction for the \try this\ problem(s) below.
try this video for example 1 - reflecting in horizontal and vertical lines
graph △abc with vertices a(1, 3), b(5, 2), and c(2, 1) and its image after the reflection described.

1. in the line n: x = -1
2. in the line m: y = 3

Explanation:

Step1: Recall Reflection Rule for Vertical Line

To reflect a point \((x,y)\) over the vertical line \(x = a\), the formula is \((x', y')=(2a - x, y)\). Here, \(a=-1\), so the rule becomes \((x', y')=(-2 - x, y)\).

Step2: Reflect Point A(1,3)

For \(A(1,3)\): \(x'=-2 - 1=-3\), \(y' = 3\). So \(A'(-3,3)\).

Step3: Reflect Point B(5,2)

For \(B(5,2)\): \(x'=-2 - 5=-7\), \(y' = 2\). So \(B'(-7,2)\).

Step4: Reflect Point C(2,1)

For \(C(2,1)\): \(x'=-2 - 2=-4\), \(y' = 1\). So \(C'(-4,1)\).

Step5: Graph Original and Image

Plot \(A(1,3)\), \(B(5,2)\), \(C(2,1)\) for \(\triangle ABC\). Then plot \(A'(-3,3)\), \(B'(-7,2)\), \(C'(-4,1)\) for the reflected triangle.

(For part 2, reflection over \(y = 3\): The rule for reflecting \((x,y)\) over \(y = b\) is \((x', y')=(x, 2b - y)\). For \(A(1,3)\): \(y'=2(3)-3 = 3\) (so \(A'(1,3)\)? Wait, no: \(A(1,3)\) is on \(y = 3\), so it maps to itself. \(B(5,2)\): \(y'=2(3)-2 = 4\), so \(B'(5,4)\). \(C(2,1)\): \(y'=2(3)-1 = 5\), so \(C'(2,5)\). Then graph these points.)

Answer:

1. Reflected vertices: \(A'(-3,3)\), \(B'(-7,2)\), \(C'(-4,1)\) (graph by plotting these and original points).
2. Reflected vertices: \(A'(1,3)\), \(B'(5,4)\), \(C'(2,5)\) (graph by plotting these and original points).