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 $(x',y')=(kx,ky)$ where $k = \frac{1}{3}$.

Step2: Identify vertices of the original figure

Let's assume the vertices of the original figure are $E(x_E,y_E)$, $F(x_F,y_F)$, $G(x_G,y_G)$, $H(x_H,y_H)$. For example, if $E(- 3,-6)$, $F(3,-6)$, $G(3,4)$, $H(0,6)$.

Step3: Calculate the new vertices

For point $E$: $x_E'=\frac{1}{3}x_E=\frac{1}{3}\times(-3)=-1$ and $y_E'=\frac{1}{3}y_E=\frac{1}{3}\times(-6)=-2$.
For point $F$: $x_F'=\frac{1}{3}x_F=\frac{1}{3}\times3 = 1$ and $y_F'=\frac{1}{3}y_F=\frac{1}{3}\times(-6)=-2$.
For point $G$: $x_G'=\frac{1}{3}x_G=\frac{1}{3}\times3 = 1$ and $y_G'=\frac{1}{3}y_G=\frac{1}{3}\times4=\frac{4}{3}$.
For point $H$: $x_H'=\frac{1}{3}x_H=\frac{1}{3}\times0 = 0$ and $y_H'=\frac{1}{3}y_H=\frac{1}{3}\times6 = 2$.

Step4: Plot the new figure

Plot the points $E'(-1,-2)$, $F'(1,-2)$, $G'(1,\frac{4}{3})$, $H'(0,2)$ and connect them to form the dilated figure.

Answer:

Plot the points obtained by multiplying the coordinates of the original - figure's vertices by $\frac{1}{3}$ and connect them to form the dilated figure.