Question

1. rhombus wxyz with vertices w(2, -4), x(6, -2), y(10, -4), and z(6, -6); k = 3/2

w (__, __)
x (__, __)
y (__, __)
z (__, __)

1. trapezoid defg with vertices d(-5, 15), e(10, 10), f(10, 5), and g(-5, 0); k = 1/5

d (__, __)
e (__, __)
f (__, __)
g (__, __)
© gina wilson (all things algebra), 2015

Explanation:

Step1: Recall dilation formula

For a point $(x,y)$ dilated by a scale - factor $k$ with the origin as the center of dilation, the new point $(x',y')$ is given by $(x',y')=(k\cdot x,k\cdot y)$.

Step2: Find the new coordinates for the rhombus

For point $W(2, - 4)$ with $k = \frac{3}{2}$

$x'_W=\frac{3}{2}\times2 = 3$, $y'_W=\frac{3}{2}\times(-4)=-6$. So $W'(3,-6)$.

For point $X(6,-2)$ with $k=\frac{3}{2}$

$x'_X=\frac{3}{2}\times6 = 9$, $y'_X=\frac{3}{2}\times(-2)=-3$. So $X'(9,-3)$.

For point $Y(10,-4)$ with $k=\frac{3}{2}$

$x'_Y=\frac{3}{2}\times10 = 15$, $y'_Y=\frac{3}{2}\times(-4)=-6$. So $Y'(15,-6)$.

For point $Z(6,-6)$ with $k=\frac{3}{2}$

$x'_Z=\frac{3}{2}\times6 = 9$, $y'_Z=\frac{3}{2}\times(-6)=-9$. So $Z'(9,-9)$.

Step3: Find the new coordinates for the trapezoid

For point $D(-5,15)$ with $k=\frac{1}{5}$

$x'_D=\frac{1}{5}\times(-5)=-1$, $y'_D=\frac{1}{5}\times15 = 3$. So $D'(-1,3)$.

For point $E(10,10)$ with $k=\frac{1}{5}$

$x'_E=\frac{1}{5}\times10 = 2$, $y'_E=\frac{1}{5}\times10 = 2$. So $E'(2,2)$.

For point $F(10,5)$ with $k=\frac{1}{5}$

$x'_F=\frac{1}{5}\times10 = 2$, $y'_F=\frac{1}{5}\times5 = 1$. So $F'(2,1)$.

For point $G(-5,0)$ with $k=\frac{1}{5}$

$x'_G=\frac{1}{5}\times(-5)=-1$, $y'_G=\frac{1}{5}\times0 = 0$. So $G'(-1,0)$.

Answer:

$W'(3,-6)$
$X'(9,-3)$
$Y'(15,-6)$
$Z'(9,-9)$
$D'(-1,3)$
$E'(2,2)$
$F'(2,1)$
$G'(-1,0)$