Question

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?
Please explain how to find the answer. Thank you!

Answer 1

To obtain the center of dilation we use the formula:
IOA'I/IOAI=IfI
this can be written as:
IOA'I=IOAIIfI
where O is the center of dilation; suppose our center is (x,y) thus plugging our values we get:
√(5-x)²+(5-y)²=2[√(0-x)²+(0-y)²]
√(5-x)²+(5-y)²=2√(x²+y²)
squaring both sides we get:
(5-x)²+(5-y)²=4(x²+y²)
to solve the above we equate as follows:
(5-x)²=4x²
x=-5 or 5/3
also
(5-y)²=4y²
y=-5 or 5/3
thus the center of dilation is:
(-5,-5)

Answer 2

The center of the dilation is (-5,-5), since it was the opposite of