Question

dilations on the coordinate plane
score: 0/3 penalty: none
question
the figure below is dilated by a factor of $\frac{1}{2}$ centered at the origin. plot the resulting image.
click twice to plot a segment.
click a segment to delete it.

Explanation:

Step1: Recall dilation formula

For a dilation centered at the origin with scale - factor $k$, if a point $(x,y)$ is dilated, the new point $(x',y')$ is given by $(x',y')=k(x,y)$. Here $k = \frac{1}{2}$.

Step2: Identify original points

Let's assume the vertices of the original quadrilateral are $S(x_1,y_1)$, $T(x_2,y_2)$, $U(x_3,y_3)$, $V(x_4,y_4)$. From the graph, if we assume $S(- 6,-4)$, $T(-2,-6)$, $U(8,6)$, $V(0,6)$.

Step3: Calculate new points

For point $S$: $x'_1=\frac{1}{2}x_1=\frac{1}{2}\times(-6)=-3$, $y'_1=\frac{1}{2}y_1=\frac{1}{2}\times(-4)=-2$.
For point $T$: $x'_2=\frac{1}{2}x_2=\frac{1}{2}\times(-2)=-1$, $y'_2=\frac{1}{2}y_2=\frac{1}{2}\times(-6)=-3$.
For point $U$: $x'_3=\frac{1}{2}x_3=\frac{1}{2}\times8 = 4$, $y'_3=\frac{1}{2}y_3=\frac{1}{2}\times6 = 3$.
For point $V$: $x'_4=\frac{1}{2}x_4=\frac{1}{2}\times0 = 0$, $y'_4=\frac{1}{2}y_4=\frac{1}{2}\times6 = 3$.

Step4: Plot new points

Plot the points $S'(-3,-2)$, $T'(-1,-3)$, $U'(4,3)$, $V'(0,3)$ and connect them to form the dilated figure.

Answer:

Plot the points $(-3,-2)$, $(-1,-3)$, $(4,3)$, $(0,3)$ and connect them to form the dilated quadrilateral.