Question

Quadrilateral ABCD has vertices A(–1, –2), B(–1, 3), C(4, 3) and D(4, –2). It's dilated by a factor of 2 with the center of dilation at the origin. What are the coordinates of the resulting quadrilateral A'B'C'D'?

Answer 1

Given:

Quadrilateral ABCD has vertices A(–1, –2), B(–1, 3), C(4, 3) and D(4, –2).

It's dilated by a factor of 2 with the center of dilation at the origin.

To find:

The coordinates of the resulting quadrilateral A'B'C'D'.

Solution:

The figure dilated by a factor of 2 with the center of dilation at the origin. So, the rule of dilation is

$(x,y)\to (2x,2y)$

Using the above rule, we get

$A(-1,-2)\to A'(2(-1),2(-2))=A'(-2,-4)$

$B(-1,3)\to B'(2(-1),2(3))=B'(-2,6)$

$C(4,3)\to C'(2(4),2(3))=C'(8,6)$

$D(4,-2)\to D'(2(4),2(-2))=D'(8,-4)$

Therefore, the coordinates of the resulting quadrilateral A'B'C'D' are A'(-2,-4), B'(-2,6), C'(8,6) and D'(8,-4).