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!
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)
