reflect ab with endpoints a(1,1) and b(3,4) in the line y = -3x. find the endpoints of the image.
endpoints: a ( , ), b ( , )

Question

Accepted Answer

# Explanation:
## Step1: Find the formula for reflection over the line $y = - 3x$
The formula for reflecting a point $(x,y)$ over the line $y=mx + c$ (here $c = 0$ and $m=-3$) can be derived using the following approach.
The line $y=-3x$ has a slope $m=-3$. The formula for reflection of a point $(x_0,y_0)$ over the line $y = mx$ is given by:
$$
\begin{cases}
x'=\frac{(1 - m^{2})x_0+2my_0}{1 + m^{2}}\
y'=\frac{2mx_0-(1 - m^{2})y_0}{1 + m^{2}}
\end{cases}
$$
Since $m=-3$, then $1+m^{2}=1 + 9=10$, $1 - m^{2}=1-9=-8$, and $2m=-6$

## Step2: Reflect point $A(1,1)$
For point $A(1,1)$:
$$
\begin{align*}
x_A'&=\frac{(1 - (-3)^{2})\times1+2\times(-3)\times1}{1+(-3)^{2}}=\frac{-8 - 6}{10}=\frac{-14}{10}=-\frac{7}{5}=-1.4\
y_A'&=\frac{2\times(-3)\times1-(1 - (-3)^{2})\times1}{1+(-3)^{2}}=\frac{-6+8}{10}=\frac{2}{10}=0.2
\end{align*}
$$

## Step3: Reflect point $B(3,4)$
For point $B(3,4)$:
$$
\begin{align*}
x_B'&=\frac{(1 - (-3)^{2})\times3+2\times(-3)\times4}{1+(-3)^{2}}=\frac{-8\times3-24}{10}=\frac{-24 - 24}{10}=\frac{-48}{10}=-4.8\
y_B'&=\frac{2\times(-3)\times3-(1 - (-3)^{2})\times4}{1+(-3)^{2}}=\frac{-18 + 32}{10}=\frac{14}{10}=1.4
\end{align*}
$$

# Answer:
$A'(-1.4,0.2)$, $B'(-4.8,1.4)$

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QUESTION IMAGE

Question

Explanation:

Step1: Find the formula for reflection over the line \(y = - 3x\)

Step2: Reflect point \(A(1,1)\)

Step3: Reflect point \(B(3,4)\)

Answer: