Question

reflect the figure over the line $y = -\frac{1}{3}x - 1$. plot all of the points of the reflected figure. you may click a plotted point to delete it. answer attempt 1 out of 3

Explanation:

Step1: Recall reflection formula

For a point $(x_0,y_0)$ reflected over the line $y = mx + c$, the general - formula for the reflected point $(x_1,y_1)$ is given by:

1. First, find the slope of the perpendicular line. The slope of the given line is $m=-\frac{1}{3}$, so the slope of the perpendicular line is $m_{\perp}=3$.
2. The equation of the line passing through $(x_0,y_0)$ with slope $m_{\perp}=3$ is $y - y_0=3(x - x_0)$, or $y=3x+(y_0 - 3x_0)$.
3. Then, find the intersection point of $y = mx + c=-\frac{1}{3}x - 1$ and $y=3x+(y_0 - 3x_0)$. Set $-\frac{1}{3}x - 1=3x+(y_0 - 3x_0)$.

\[-\frac{1}{3}x-3x=y_0 - 3x_0 + 1\]
\[-\frac{1 + 9}{3}x=y_0 - 3x_0 + 1\]
\[x=\frac{-3(y_0 - 3x_0 + 1)}{10}\]
Substitute $x$ into $y =-\frac{1}{3}x - 1$ to get $y$.

1. Another way is to use the following transformation formula:

The formula for reflecting a point $(x_0,y_0)$ over the line $y=mx + c$ is:
\[x_1=\frac{1 - m^{2}}{1 + m^{2}}x_0+\frac{2m}{1 + m^{2}}(y_0 - c)\]
\[y_1=\frac{2m}{1 + m^{2}}x_0+\frac{m^{2}-1}{1 + m^{2}}y_0+\frac{2c}{1 + m^{2}}\]
Here, $m =-\frac{1}{3}$ and $c=-1$.
\[1 + m^{2}=1+\frac{1}{9}=\frac{9 + 1}{9}=\frac{10}{9}\]
\[x_1=\frac{1-\frac{1}{9}}{\frac{10}{9}}x_0+\frac{2(-\frac{1}{3})}{\frac{10}{9}}(y_0+1)=\frac{\frac{8}{9}}{\frac{10}{9}}x_0-\frac{\frac{2}{3}}{\frac{10}{9}}(y_0 + 1)=\frac{4}{5}x_0-\frac{3}{5}(y_0 + 1)\]
\[y_1=\frac{2(-\frac{1}{3})}{\frac{10}{9}}x_0+\frac{\frac{1}{9}-1}{\frac{10}{9}}y_0+\frac{2(-1)}{\frac{10}{9}}=-\frac{3}{5}x_0-\frac{4}{5}y_0-\frac{9}{5}\]
For each vertex of the polygon (let the vertex be $(x_0,y_0)$), calculate the reflected - point $(x_1,y_1)$ using the above formula and then plot these new points to get the reflected figure.

Since the vertices of the original polygon are not given numerically, we can't give the exact numerical coordinates of the reflected points. But the general method to find the reflected points of a figure over the line $y =-\frac{1}{3}x - 1$ is as described above.

Answer:

Use the transformation formulas $x_1=\frac{4}{5}x_0-\frac{3}{5}(y_0 + 1)$ and $y_1=-\frac{3}{5}x_0-\frac{4}{5}y_0-\frac{9}{5}$ to find the coordinates of the reflected vertices of the figure, then plot the reflected figure.