Question

Triangle GHJ with vertices G(-2, 4), H(3, 6), and J(3, -2) is dilated by a factor of 1/3 centered at the origin. What are the coordinates of G' after the dilation?

Answer 1

Answer:

The coordinates of G' are $G'(x,y) = \left(-\frac{2}{3}, \frac{4}{3} \right)$.

Step-by-step explanation:

From Linear Algebra, we define the dilation of point G by the following expression:

$G'(x,y) = O(x,y) + k\cdot [G(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation, dimensionless.

$k$ - Scale factor, dimensionless.

$G(x,y)$ - Coordinates of point G, dimensionless.

$G'(x,y)$ - Coordinates of point G', dimensionless.

If we know that $O(x,y) = (0,0)$, $k = \frac{1}{3}$, $G(x,y) = (-2,4)$, then the point G' is:

$G'(x,y) = (0,0) + \frac{1}{3}\cdot [(-2,4)-(0,0)]$

$G'(x,y) = \left(-\frac{2}{3}, \frac{4}{3} \right)$

The coordinates of G' are $G'(x,y) = \left(-\frac{2}{3}, \frac{4}{3} \right)$.