Question

flect 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:

Answer:

To reflect a point $(x_0,y_0)$ over the line $y = mx + c$ (where $m=-\frac{1}{3}$ and $c = - 1$), we use the following general - formula for reflection of a point $(x_0,y_0)$ over the line $y=mx + c$:
The formula for the reflected point $(x_1,y_1)$ is given by:
\[x_1=x_0-\frac{2m(mx_0 - y_0 + c)}{1 + m^{2}}\]
\[y_1=y_0+\frac{2(mx_0 - y_0 + c)}{1 + m^{2}}\]

1. First, identify the slope $m$ and the y - intercept $c$ of the line of reflection:

• Here, $m =-\frac{1}{3}$ and $c=-1$.

1. Let's take a general point $(x_0,y_0)$ on the original figure.

• Calculate $x_1$:
• Substitute $m =-\frac{1}{3}$ into the formula for $x_1$:
• $x_1=x_0-\frac{2(-\frac{1}{3})(-\frac{1}{3}x_0 - y_0-1)}{1+(-\frac{1}{3})^{2}}$
• First, simplify the denominator: $1+(-\frac{1}{3})^{2}=1+\frac{1}{9}=\frac{9 + 1}{9}=\frac{10}{9}$.
• Then, simplify the numerator: $2(-\frac{1}{3})(-\frac{1}{3}x_0 - y_0 - 1)=-\frac{2}{3}(-\frac{1}{3}x_0 - y_0 - 1)=\frac{2}{9}x_0+\frac{2}{3}y_0+\frac{2}{3}$.
• So, $x_1=x_0-\frac{\frac{2}{9}x_0+\frac{2}{3}y_0+\frac{2}{3}}{\frac{10}{9}}=x_0-\frac{9}{10}(\frac{2}{9}x_0+\frac{2}{3}y_0+\frac{2}{3})$.
• Expand the right - hand side: $x_0-\frac{1}{5}x_0-\frac{3}{5}y_0-\frac{3}{5}=\frac{4}{5}x_0-\frac{3}{5}y_0-\frac{3}{5}$.
• Calculate $y_1$:
• Substitute $m =-\frac{1}{3}$ into the formula for $y_1$:
• $y_1=y_0+\frac{2(-\frac{1}{3}x_0 - y_0 - 1)}{\frac{10}{9}}$.
• First, simplify the numerator: $2(-\frac{1}{3}x_0 - y_0 - 1)=-\frac{2}{3}x_0-2y_0 - 2$.
• Then, $y_1=y_0+\frac{9}{10}(-\frac{2}{3}x_0-2y_0 - 2)$.
• Expand the right - hand side: $y_1=y_0-\frac{3}{5}x_0-\frac{9}{5}y_0-\frac{9}{5}=-\frac{3}{5}x_0-\frac{4}{5}y_0-\frac{9}{5}$.

1. For each vertex $(x_0,y_0)$ of the original polygon:

• Calculate the corresponding reflected vertex $(x_1,y_1)$ using the above formulas.
• For example, if one of the vertices of the original polygon is $(x_0,y_0)=( - 4,-4)$:
• $x_1=\frac{4}{5}(-4)-\frac{3}{5}(-4)-\frac{3}{5}=\frac{-16 + 12-3}{5}=-\frac{7}{5}=-1.4$.
• $y_1=-\frac{3}{5}(-4)-\frac{4}{5}(-4)-\frac{9}{5}=\frac{12 + 16-9}{5}=\frac{19}{5}=3.8$.
• Repeat this process for all vertices of the original figure and then plot the new points to get the reflected figure.

Since we don't have the exact coordinates of the vertices of the polygon in the figure, the general method to find the reflected points is as above. If we had the coordinates of the vertices of the original polygon (say $(x_1,y_1),(x_2,y_2),\cdots$), we would substitute each pair into the formulas for $(x_1,y_1)$ of the reflected point to get the new set of vertices for the reflected polygon and then plot them.