Question

the figure below is dilated by a factor of $\frac{1}{3}$ 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

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

Step2: Identify original points

Let's assume the coordinates of the vertices of the polygon are: $V(0, - 6)$, $W(6,-3)$, $X(6,3)$, $Y(4,6)$, $Z(-9,6)$.

Step3: Calculate new coordinates

For point $V(0,-6)$:
$x'=\frac{1}{3}\times0 = 0$, $y'=\frac{1}{3}\times(-6)=-2$. So the new point $V'(0, - 2)$.
For point $W(6,-3)$:
$x'=\frac{1}{3}\times6 = 2$, $y'=\frac{1}{3}\times(-3)=-1$. So the new point $W'(2,-1)$.
For point $X(6,3)$:
$x'=\frac{1}{3}\times6 = 2$, $y'=\frac{1}{3}\times3 = 1$. So the new point $X'(2,1)$.
For point $Y(4,6)$:
$x'=\frac{1}{3}\times4=\frac{4}{3}$, $y'=\frac{1}{3}\times6 = 2$. So the new point $Y'(\frac{4}{3},2)$.
For point $Z(-9,6)$:
$x'=\frac{1}{3}\times(-9)=-3$, $y'=\frac{1}{3}\times6 = 2$. So the new point $Z'(-3,2)$.

Step4: Plot new points

Plot the points $V'(0, - 2)$, $W'(2,-1)$, $X'(2,1)$, $Y'(\frac{4}{3},2)$, $Z'(-3,2)$ on the coordinate - plane and connect them in the same order as the original polygon.

Answer:

Plot the points $V'(0, - 2)$, $W'(2,-1)$, $X'(2,1)$, $Y'(\frac{4}{3},2)$, $Z'(-3,2)$ and connect them.