If square ABCD is defined by the coordinates A(0, 0), B(0, -5), C(-5, -5), D(-5, 0) is dilated by a scale factor of 2, with resulting vertex A' at (5, 5). What is the center of the dilation?

Question

A. (0,0)

B. (5,5)

C. (-5,0)

D. (-5,-5)

Answer 1

Let the centre be at O(p,q). Distance x of A from O is 0-p=-p and distance y of A from O is 0-q=-q.

Now apply the dilation of 2, so distance x of A' from O is -2p and distance y is -2q.

We add these displacements to O(p,q) to find the new coords of A': p+(-2p)=-p and q+(-2q)=-q.

So we have A'(-p,-q)=(5,5). Therefore p=-5 and q=-5 and the centre is O(-5,-5), coincident with vertex C and answer choice D.