Question

the figure below is dilated by a factor of 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 rule

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 $(x',y')=(k x,k y)$. Here $k = \frac{1}{2}$.

Step2: Identify original points

Let's assume the coordinates of the vertices of the polygon are: Let $U(x_U,y_U)$, $T(x_T,y_T)$, $S(x_S,y_S)$, $R(x_R,y_R)$, $V(x_V,y_V)$. Suppose $U(- 6,6)$, $T(6,2)$, $S(4, - 4)$, $R(6,-6)$, $V(-9,0)$.

Step3: Calculate new coordinates

For point $U$: $x_U'=\frac{1}{2}x_U=\frac{1}{2}\times(-6)=-3$, $y_U'=\frac{1}{2}y_U=\frac{1}{2}\times6 = 3$.
For point $T$: $x_T'=\frac{1}{2}x_T=\frac{1}{2}\times6 = 3$, $y_T'=\frac{1}{2}y_T=\frac{1}{2}\times2 = 1$.
For point $S$: $x_S'=\frac{1}{2}x_S=\frac{1}{2}\times4 = 2$, $y_S'=\frac{1}{2}y_S=\frac{1}{2}\times(-4)=-2$.
For point $R$: $x_R'=\frac{1}{2}x_R=\frac{1}{2}\times6 = 3$, $y_R'=\frac{1}{2}y_R=\frac{1}{2}\times(-6)=-3$.
For point $V$: $x_V'=\frac{1}{2}x_V=\frac{1}{2}\times(-9)=-\frac{9}{2}=-4.5$, $y_V'=\frac{1}{2}y_V=\frac{1}{2}\times0 = 0$.

Step4: Plot new points

Plot the points $U'(-3,3)$, $T'(3,1)$, $S'(2,-2)$, $R'(3,-3)$, $V'(-4.5,0)$ on the coordinate - plane and connect them in the same order as the original polygon to get the dilated image.

Answer:

Plot the points $(-3,3)$, $(3,1)$, $(2,-2)$, $(3,-3)$, $(-4.5,0)$ and connect them to form the dilated figure.