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April 17, 2014

Posted by **Cindy** on Monday, December 19, 2011 at 3:50pm.

- geometry -
**Steve**, Monday, December 19, 2011 at 6:42pmI 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.

- geometry -
**Reiny**, Monday, December 19, 2011 at 7:22pmDid 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)

or

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

- Note to Steve -
**Reiny**, Monday, December 19, 2011 at 7:28pmSeems 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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