Question

the figure below is dilated by a factor of 1/3 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

If a point $(x,y)$ is dilated by a factor $k$ centered at the origin, the new - point $(x',y')$ is given by $(x',y')=(k x,k y)$. Here $k = \frac{1}{3}$.

Step2: Identify original points

Let's assume the coordinates of the vertices of the polygon are $Z(x_Z,y_Z)$, $Y(x_Y,y_Y)$, $X(x_X,y_X)$, $W(x_W,y_W)$, $V(x_V,y_V)$.

Step3: Calculate new coordinates

For a vertex with coordinates $(x,y)$ of the original figure, the new coordinates after dilation are $x'=\frac{1}{3}x$ and $y'=\frac{1}{3}y$. For example, if $Z$ has coordinates $(- 8,6)$, then $Z'$ has coordinates $(\frac{1}{3}\times(-8),\frac{1}{3}\times6)=(-\frac{8}{3},2)$. Do this for all vertices of the polygon.

Step4: Plot new - points

Plot the new - points $Z'$, $Y'$, $X'$, $W'$, $V'$ on the coordinate plane and connect them in the same order as the original polygon to get the dilated image.

Since we don't have the actual coordinates of the vertices given in text, the general procedure for dilation is as above. If you provide the coordinates of the vertices of the polygon (e.g., $Z(-8,6)$, $Y(6,6)$, $X(6,3)$, $W(6, - 3)$, $V(0,-6)$), we can calculate the exact new coordinates:
Let's assume $Z(-8,6)$, $Y(6,6)$, $X(6,3)$, $W(6,-3)$, $V(0,-6)$
For $Z$: $Z'=(\frac{1}{3}\times(-8),\frac{1}{3}\times6)=(-\frac{8}{3},2)$
For $Y$: $Y'=(\frac{1}{3}\times6,\frac{1}{3}\times6)=(2,2)$
For $X$: $X'=(\frac{1}{3}\times6,\frac{1}{3}\times3)=(2,1)$
For $W$: $W'=(\frac{1}{3}\times6,\frac{1}{3}\times(-3))=(2, - 1)$
For $V$: $V'=(\frac{1}{3}\times0,\frac{1}{3}\times(-6))=(0,-2)$

Answer:

Plot the points $(-\frac{8}{3},2)$, $(2,2)$, $(2,1)$, $(2,-1)$, $(0,-2)$ and connect them in order to get the dilated image.