If you dilate triangle XYZ to the form X′Y′Z′ using a magnitude of 2 and a center of dilation at (2,0) , what are the coordinates of the dilated triangle? X(2,4) Z(7,2) Y(6,0)
To dilate the triangle with a magnitude of 2 and a center of dilation at (2,0), we will multiply the distance between each vertex and the center of dilation by 2 and add the result to the coordinates of the center of dilation to obtain the coordinates of the dilated triangle.
For vertex X(2,4):
The distance between X and the center of dilation is:
ΔX = 2 - 2 = 0 (change in x-coordinate)
ΔY = 4 - 0 = 4 (change in y-coordinate)
To dilate this distance by a magnitude of 2, we multiply ΔX and ΔY by 2:
ΔX' = 2 * 0 = 0
ΔY' = 2 * 4 = 8
Finally, add these changes to the coordinates of the center of dilation (2,0):
X' = (2 + ΔX', 0 + ΔY') = (2 + 0, 0 + 8) = (2, 8)
Similarly, for vertex Z(7,2):
ΔX = 7 - 2 = 5
ΔY = 2 - 0 = 2
ΔX' = 2 * 5 = 10
ΔY' = 2 * 2 = 4
Z' = (2 + ΔX', 0 + ΔY') = (2 + 10, 0 + 4) = (12, 4)
And for vertex Y(6,0):
ΔX = 6 - 2 = 4
ΔY = 0 - 0 = 0
ΔX' = 2 * 4 = 8
ΔY' = 2 * 0 = 0
Y' = (2 + ΔX', 0 + ΔY') = (2 + 8, 0 + 0) = (10, 0)
Therefore, the coordinates of the dilated triangle X'Y'Z' are X'(2,8), Z'(12,4), Y'(10,0).