I had to puzzle this out myself. In this case, we need to
(a) find the center of dilation
(b) verify that the dilation of A to A' is the same as that of B to B'
Since dilation is a linear scaling, both A and B are moving toward some point C. Naturally, if the dilation were 0, both points would contract to the same point: the intersection of the lines AA' and BB'
The line containing A and A' is (y-3)/(x-6) = 1/3
The line containing B and B' is (y+3)/(x-3) = -1/2
These lines intersect at (-3,0)
So, now, let's verify that the scale factor is the same along both directions
CA'/CA = √90/√160 = 3/4
CB'/CB = √45/√80 = 3/4
So, the scale factor is 3/4, using (-3,0) as the center of dilation.
Did you notice that slope AB = slope A'B' = 2 ?
So by extending AA' and BB' until they meet we can find the centre of dilation
Making a neat graph shows that this centre of dilation is C(-3,0)
You can find the equation of AA',which was y = (1/3)x + 1, and the equation of BB', which was y = (-1/2)x - 3/2
solving these two to get (-3,0)
AC = √(144+16) = √160 = 4√10
A'C = √(81+9) = √90 = 3√10
A'C/AC = 3V10/(4√10) = 3/4
BC = √(64+16) = √80 = 4√5
B'C = √(36+9) = √45 = 3√5
B'C/BC = 3√5/(4√5) = 3/4
scale factor is 3/4
Seems like we were on some kind of Vulcan mind-meld.
Even our choice of C for the centre of dilation was the same, spooky!
I had never seen that kind of question, and actually printed myself out a sheet of graph paper, lol
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