Question

graph the image of △stu after a dilation with a scale factor of $\frac{1}{4}$, centered at the origin.

Explanation:

Step1: Recall dilation formula

If a point $(x,y)$ is dilated with a scale - 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: Assume coordinates of vertices

Let's assume the coordinates of the vertices of $\triangle STU$ are $S(x_1,y_1)$, $T(x_2,y_2)$ and $U(x_3,y_3)$. From the graph, if we assume $S(- 6,-8)$, $T(8,-8)$ and $U(-8,4)$.

Step3: Calculate new coordinates for $S$

For point $S(-6,-8)$:
$x'_1=\frac{1}{4}\times(-6)=-\frac{6}{4}=-\frac{3}{2}$
$y'_1=\frac{1}{4}\times(-8)= - 2$
So the new point $S'$ is $(-\frac{3}{2},-2)$.

Step4: Calculate new coordinates for $T$

For point $T(8,-8)$:
$x'_2=\frac{1}{4}\times8 = 2$
$y'_2=\frac{1}{4}\times(-8)=-2$
So the new point $T'$ is $(2,-2)$.

Step5: Calculate new coordinates for $U$

For point $U(-8,4)$:
$x'_3=\frac{1}{4}\times(-8)=-2$
$y'_3=\frac{1}{4}\times4 = 1$
So the new point $U'$ is $(-2,1)$.

Step6: Graph the new triangle

Plot the points $S'(-\frac{3}{2},-2)$, $T'(2,-2)$ and $U'(-2,1)$ on the same coordinate - plane and connect them to form the dilated triangle $\triangle S'T'U'$.

Answer:

Graph the triangle with vertices $(-\frac{3}{2},-2)$, $(2,-2)$ and $(-2,1)$.