Question

for exercises 1–5, use the diagram.

1. what composition of two rigid motions maps

△abc to △a′b′c′?
t < -5,0> t x-axis
for exercises 2–5, find the coordinates
of p′ under each transformation. suppose the
equation of line m is y = 2 and the equation of line n is x = -1.

1. ( t_{langle -2, 0

angle} circ r_m ) p (1,1) >(1,3)> p (-1,3)

1. ( t_{langle 0, -5

angle} circ r_n ) p(1,1)> p(-3,1)> p(-3,-4)

1. ( t_{langle 0, 2

angle} circ r_{y\text{-axis}} ) (-1,3)

1. ( t_{langle 3, 0

angle} circ r_{x\text{-axis}} ) (4,-1)
for exercises 6–12, write a rigid motion that
produces each image.

1. ( \triangle abc \to \triangle def ) translation
2. ( \triangle abc \to \triangle ghj )
3. ( \triangle abc \to \triangle klm )
4. ( \triangle abc \to \triangle npq )
5. ( \triangle abc \to \triangle rst )
6. ( \triangle def \to \triangle ghj )
7. ( \triangle ghj \to \triangle klm )
8. understand define the term slide reflection

Explanation:

Step1: Analyze the transformation \( T_{\langle -2, 0

angle} \circ r_m \)
First, apply the reflection \( r_m \) (over line \( y = 2 \)) to point \( P(1,1) \). The distance from \( P \)'s \( y \)-coordinate (1) to the line \( y = 2 \) is \( 2 - 1 = 1 \). So, reflecting over \( y = 2 \), the new \( y \)-coordinate is \( 2 + 1 = 3 \), and the \( x \)-coordinate remains 1. So, \( r_m(P) = (1, 3) \).

Step2: Apply the translation \( T_{\langle -2, 0

angle} \)
A translation \( T_{\langle -2, 0
angle} \) means we subtract 2 from the \( x \)-coordinate and keep the \( y \)-coordinate the same. So, for the point \( (1, 3) \), the new \( x \)-coordinate is \( 1 - 2 = -1 \), and the \( y \)-coordinate is 3. Thus, \( T_{\langle -2, 0
angle}(1, 3) = (-1, 3) \).

Answer:

The coordinates of \( P' \) are \((-1, 3)\)