Question

If rectangle ABCD, defined by the coordinates A(0, 0), B(0, -4), C(-2, -4), D(-2, 0), is dilated by a scale factor of 2, with resulting vertex A' at (2, 4), what is the center of the dilation? A) (0, 0) Eliminate B) (2, 4) C) (0, -4) D) (-2, -4)

Answer 1

Answer:

D) (-2, -4).

Step-by-step explanation:

Let $(x, y)$ be the center of dilation. Imagine that $(x, y)$ is the center of origin of a new Cartesian Plane. What would the coordinates of point A and A' on the new plane?

• A: $(0 - x, 0 - y)$ ⇒ $(-x, -y)$.
• A': $(2 - x, 4 - y)$.

Dilating $(-x, -y)$ about the "origin" of the new plane by a factor of two will give the point $(-{\bf 2}x, -{\bf 2}y)$, also on the new plane.

For coordinates of A' on the original plane,

$(-{2}x, -{2}y) = (2 - x, 4 - y)$.

As a result,

$\left \{ \begin{array}{l}-2x = 2 - x\\-2 y = 4 - y\end{array}$.

$\left \{ \begin{array}{l}-x = 2\\- y = 4\end{array}$.

$\left \{ \begin{array}{l}x = -2\\y = -4\end{array}$.

In other words, ${\bf (-2, -4)}$ is the center of dilation.