Question

dilations on the coordinate plane
score: 2/3 penalty: none
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 rule

For a dilation of factor $k = \frac{1}{3}$ centered at the origin, if a point has coordinates $(x,y)$, the new - coordinates $(x',y')$ are given by $(x',y')=(k x,k y)$.

Step2: Find new coordinates of point Z

Suppose $Z=(- 9,6)$. Then $x=-9,y = 6$. So $x'=\frac{1}{3}\times(-9)=-3$ and $y'=\frac{1}{3}\times6 = 2$. The new point $Z'$ is $(-3,2)$.

Step3: Find new coordinates of point Y

Suppose $Y=(5,6)$. Then $x = 5,y = 6$. So $x'=\frac{1}{3}\times5=\frac{5}{3}$ and $y'=\frac{1}{3}\times6 = 2$. The new point $Y'$ is $(\frac{5}{3},2)$.

Step4: Find new coordinates of point X

Suppose $X=(6,3)$. Then $x = 6,y = 3$. So $x'=\frac{1}{3}\times6 = 2$ and $y'=\frac{1}{3}\times3=1$. The new point $X'$ is $(2,1)$.

Step5: Find new coordinates of point W

Suppose $W=(6,-3)$. Then $x = 6,y=-3$. So $x'=\frac{1}{3}\times6 = 2$ and $y'=\frac{1}{3}\times(-3)=-1$. The new point $W'$ is $(2,-1)$.

Step6: Find new coordinates of point V

Suppose $V=(0,-6)$. Then $x = 0,y=-6$. So $x'=\frac{1}{3}\times0 = 0$ and $y'=\frac{1}{3}\times(-6)=-2$. The new point $V'$ is $(0,-2)$.

Step7: Plot new points

Plot the points $Z'(-3,2)$, $Y'(\frac{5}{3},2)$, $X'(2,1)$, $W'(2,-1)$ and $V'(0,-2)$ and connect them in the same order as the original figure.

Answer:

Plot the points $(-3,2)$, $(\frac{5}{3},2)$, $(2,1)$, $(2,-1)$, $(0,-2)$ and connect them in order.