Question

the figure below is dilated by a factor of 1/4 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 $k = \frac{1}{4}$.

Step2: Identify original points

Let's assume the vertices of the triangle are $J(0,8)$, $K(0,0)$, and $I(6, - 4)$, $G(-2,-4)$.

Step3: Calculate new points for $J$

For point $J(0,8)$, $x = 0$, $y = 8$, and $k=\frac{1}{4}$. Then $x'=k x=\frac{1}{4}\times0 = 0$ and $y'=k y=\frac{1}{4}\times8 = 2$. So the new point $J'$ is $(0,2)$.

Step4: Calculate new points for $K$

For point $K(0,0)$, $x = 0$, $y = 0$, and $k=\frac{1}{4}$. Then $x'=k x=\frac{1}{4}\times0 = 0$ and $y'=k y=\frac{1}{4}\times0 = 0$. So the new point $K'$ is $(0,0)$.

Step5: Calculate new points for $I$

For point $I(6,-4)$, $x = 6$, $y=-4$, and $k = \frac{1}{4}$. Then $x'=k x=\frac{1}{4}\times6=\frac{3}{2}$ and $y'=k y=\frac{1}{4}\times(-4)=-1$. So the new point $I'$ is $(\frac{3}{2},-1)$.

Step6: Calculate new points for $G$

For point $G(-2,-4)$, $x=-2$, $y = - 4$, and $k=\frac{1}{4}$. Then $x'=k x=\frac{1}{4}\times(-2)=-\frac{1}{2}$ and $y'=k y=\frac{1}{4}\times(-4)=-1$. So the new point $G'$ is $(-\frac{1}{2},-1)$.

Step7: Plot the new points

Plot the points $J'(0,2)$, $K'(0,0)$, $I'(\frac{3}{2},-1)$, and $G'(-\frac{1}{2},-1)$ on the coordinate - plane and connect them to form the dilated triangle.

Answer:

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