Question

1. square abcd has coordinates a(4, 4), b(8, 4), c(8, 0), and d(4, 0). a dilation of square abcd has coordinates a(0, 0), b(2, 0), c(2, -2), and d(0, -2). what is the center of dilation? center of dilation:

Explanation:

Step1: Recall dilation formula

Let the center of dilation be $(x_0,y_0)$ and a point $(x,y)$ be dilated to $(x',y')$. The formula for dilation is $x'=k(x - x_0)+x_0$ and $y'=k(y - y_0)+y_0$. We can also use the fact that the center of dilation lies on the lines connecting corresponding points.
Let's consider two - pairs of corresponding points, say $A(4,4)$ and $A'(0,0)$, and $B(8,4)$ and $B'(2,0)$.
The equation of the line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is $y - y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
The equation of the line passing through $A(4,4)$ and $A'(0,0)$ is $y=\frac{4 - 0}{4 - 0}x=x$.

Step2: Find the equation of the line through another pair

The equation of the line passing through $B(8,4)$ and $B'(2,0)$:
The slope $m=\frac{4 - 0}{8 - 2}=\frac{2}{3}$. Using the point - slope form with the point $B'(2,0)$, we have $y-0=\frac{2}{3}(x - 2)$, so $y=\frac{2}{3}x-\frac{4}{3}$.

Step3: Solve the system of equations

We need to solve the system of equations

$$\begin{cases}y = x\\y=\frac{2}{3}x-\frac{4}{3}\end{cases}$$

Substitute $y = x$ into $y=\frac{2}{3}x-\frac{4}{3}$:
$x=\frac{2}{3}x-\frac{4}{3}$
$x-\frac{2}{3}x=-\frac{4}{3}$
$\frac{1}{3}x=-\frac{4}{3}$
$x = 6$
Since $y = x$, then $y = 2$.
So the center of dilation is $(6,2)$.

Answer:

$(6,2)$