Question

the figure below is dilated by a factor of $\frac{1}{2}$ 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 $k = \frac{1}{2}$.

Step2: Identify original points

Let's assume the coordinates of the vertices of the polygon are $U(x_U,y_U)$, $T(x_T,y_T)$, $S(x_S,y_S)$, $R(x_R,y_R)$ and $V(x_V,y_V)$. For example, if $U=(- 4,6)$, $T=(6,2)$, $S=(4, - 4)$, $R=(2,-6)$ and $V=(-8,0)$.

Step3: Calculate new coordinates

For point $U$: $x_U'=\frac{1}{2}x_U=\frac{1}{2}\times(-4)=-2$, $y_U'=\frac{1}{2}y_U=\frac{1}{2}\times6 = 3$.
For point $T$: $x_T'=\frac{1}{2}x_T=\frac{1}{2}\times6 = 3$, $y_T'=\frac{1}{2}y_T=\frac{1}{2}\times2 = 1$.
For point $S$: $x_S'=\frac{1}{2}x_S=\frac{1}{2}\times4 = 2$, $y_S'=\frac{1}{2}y_S=\frac{1}{2}\times(-4)=-2$.
For point $R$: $x_R'=\frac{1}{2}x_R=\frac{1}{2}\times2 = 1$, $y_R'=\frac{1}{2}y_R=\frac{1}{2}\times(-6)=-3$.
For point $V$: $x_V'=\frac{1}{2}x_V=\frac{1}{2}\times(-8)=-4$, $y_V'=\frac{1}{2}y_V=\frac{1}{2}\times0 = 0$.

Step4: Plot new points

Plot the points $U'(-2,3)$, $T'(3,1)$, $S'(2,-2)$, $R'(1,-3)$ and $V'(-4,0)$ and connect them in the same order as the original polygon to get the dilated figure.

Since this is a plotting task, no single - numerical answer can be provided in a traditional sense. The key is to calculate the new coordinates of the vertices of the figure after dilation and then plot them on the coordinate plane.

Answer:

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 $k = \frac{1}{2}$.

Step2: Identify original points

Let's assume the coordinates of the vertices of the polygon are $U(x_U,y_U)$, $T(x_T,y_T)$, $S(x_S,y_S)$, $R(x_R,y_R)$ and $V(x_V,y_V)$. For example, if $U=(- 4,6)$, $T=(6,2)$, $S=(4, - 4)$, $R=(2,-6)$ and $V=(-8,0)$.

Step3: Calculate new coordinates

For point $U$: $x_U'=\frac{1}{2}x_U=\frac{1}{2}\times(-4)=-2$, $y_U'=\frac{1}{2}y_U=\frac{1}{2}\times6 = 3$.
For point $T$: $x_T'=\frac{1}{2}x_T=\frac{1}{2}\times6 = 3$, $y_T'=\frac{1}{2}y_T=\frac{1}{2}\times2 = 1$.
For point $S$: $x_S'=\frac{1}{2}x_S=\frac{1}{2}\times4 = 2$, $y_S'=\frac{1}{2}y_S=\frac{1}{2}\times(-4)=-2$.
For point $R$: $x_R'=\frac{1}{2}x_R=\frac{1}{2}\times2 = 1$, $y_R'=\frac{1}{2}y_R=\frac{1}{2}\times(-6)=-3$.
For point $V$: $x_V'=\frac{1}{2}x_V=\frac{1}{2}\times(-8)=-4$, $y_V'=\frac{1}{2}y_V=\frac{1}{2}\times0 = 0$.

Step4: Plot new points

Plot the points $U'(-2,3)$, $T'(3,1)$, $S'(2,-2)$, $R'(1,-3)$ and $V'(-4,0)$ and connect them in the same order as the original polygon to get the dilated figure.

Since this is a plotting task, no single - numerical answer can be provided in a traditional sense. The key is to calculate the new coordinates of the vertices of the figure after dilation and then plot them on the coordinate plane.