Question

1. reflection across $y=1$

$p(2, 2), q(3, 4), r(4, 4), s(2, 0)$

extra credit:

1. reflect the triangle across the given line

$y=2x - 1$

Explanation:

For problem 9:

Step1: Recall reflection rule for $y=k$

For a point $(x,y)$, reflection across $y=1$ uses $y' = 2(1) - y$, $x'=x$.

Step2: Reflect point $P(2,2)$

$x'=2$, $y'=2(1)-2=0$ → $P'(2,0)$

Step3: Reflect point $Q(3,4)$

$x'=3$, $y'=2(1)-4=-2$ → $Q'(3,-2)$

Step4: Reflect point $R(4,4)$

$x'=4$, $y'=2(1)-4=-2$ → $R'(4,-2)$

Step5: Reflect point $S(2,0)$

$x'=2$, $y'=2(1)-0=2$ → $S'(2,2)$

Step1: Recall reflection formula for line $y=mx+b$

For a point $(x_0,y_0)$, reflection $(x',y')$ is:

$$\begin{align*} x'&=x_0 - 2m\frac{mx_0 - y_0 + b}{m^2+1}\\ y'&=y_0 + 2\frac{mx_0 - y_0 + b}{m^2+1} \end{align*}$$

Here $m=2$, $b=-1$

Step2: Calculate for $A(-4,-2)$

First compute $\frac{mx_0 - y_0 + b}{m^2+1}=\frac{2(-4)-(-2)-1}{4+1}=\frac{-8+2-1}{5}=\frac{-7}{5}$
$x'=-4 - 2(2)(\frac{-7}{5})=-4+\frac{28}{5}=\frac{8}{5}=1.6$
$y'=-2 + 2(\frac{-7}{5})=-2-\frac{14}{5}=\frac{-24}{5}=-4.8$ → $A'(\frac{8}{5}, -\frac{24}{5})$

Step3: Calculate for $B(-2,1)$

$\frac{2(-2)-(1)-1}{5}=\frac{-4-1-1}{5}=\frac{-6}{5}$
$x'=-2 - 2(2)(\frac{-6}{5})=-2+\frac{24}{5}=\frac{14}{5}=2.8$
$y'=1 + 2(\frac{-6}{5})=1-\frac{12}{5}=\frac{-7}{5}=-1.4$ → $B'(\frac{14}{5}, -\frac{7}{5})$

Step4: Calculate for $C(-5,-3)$

$\frac{2(-5)-(-3)-1}{5}=\frac{-10+3-1}{5}=\frac{-8}{5}$
$x'=-5 - 2(2)(\frac{-8}{5})=-5+\frac{32}{5}=\frac{7}{5}=1.4$
$y'=-3 + 2(\frac{-8}{5})=-3-\frac{16}{5}=\frac{-31}{5}=-6.2$ → $C'(\frac{7}{5}, -\frac{31}{5})$

Answer:

Reflected points: $P'(2,0)$, $Q'(3,-2)$, $R'(4,-2)$, $S'(2,2)$

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For extra credit problem 11:

First, identify the triangle's vertices from the graph: let's assume they are $A(-4,-2)$, $B(-2,1)$, $C(-5,-3)$ (from the grid)