Question

reflect the figure over the line y = x - 3. plot all of the points of the reflected figure. you may click a plotted point to delete it.

Explanation:

Step1: Recall reflection formula

The formula to reflect a point $(x_0,y_0)$ over the line $y = mx + c$ (here $m = 1$ and $c=-3$) is given by:
The mid - point of the line segment joining $(x_0,y_0)$ and its reflection $(x_1,y_1)$ lies on the line $y=mx + c$, and the line joining $(x_0,y_0)$ and $(x_1,y_1)$ is perpendicular to $y=mx + c$. For a line $y = mx + c$, the slope of the perpendicular line is $-\frac{1}{m}$. Since $m = 1$ for $y=x - 3$, the slope of the perpendicular line is $- 1$.
The mid - point of the points $(x_0,y_0)$ and $(x_1,y_1)$ is $(\frac{x_0 + x_1}{2},\frac{y_0 + y_1}{2})$, and it lies on the line $y=x - 3$, so $\frac{y_0 + y_1}{2}=\frac{x_0 + x_1}{2}-3$. Also, the slope of the line joining $(x_0,y_0)$ and $(x_1,y_1)$ is $\frac{y_1 - y_0}{x_1 - x_0}=-1$, which implies $y_1 - y_0=-(x_1 - x_0)$ or $y_1=-x_1+(x_0 + y_0)$.
Substitute $y_1$ into the mid - point equation:
\[

$$\begin{align*} \frac{-x_1+(x_0 + y_0)+y_0}{2}&=\frac{x_0 + x_1}{2}-3\\ -x_1+(x_0 + 2y_0)&=x_0 + x_1-6\\ -2x_1&=-6 - 2y_0\\ x_1&=y_0 + 3 \end{align*}$$

\]
And since $y_1=-x_1+(x_0 + y_0)$, substituting $x_1=y_0 + 3$ gives $y_1=x_0-3$.

Step2: Apply formula to each vertex

Let the vertices of the given figure be $(x_i,y_i)$ for $i = 1,2,3,4$. For each vertex, calculate the reflected vertex $(x_{i1},y_{i1})$ using $x_{i1}=y_i + 3$ and $y_{i1}=x_i-3$. Then plot these new points on the coordinate - plane.

Answer:

Plot the points obtained by using the transformation $(x,y)\to(y + 3,x - 3)$ for each vertex of the original figure.