Answer:
10. T(-20, 9)
11. R(-1, -9)
12. D(-6, 5)
Explanation:
The relationship between end points and the midpoint can be solved generically for one of the end points. That relation can be used for the different points in these problems.
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generic solution
The coordinates of midpoint M is related to the coordinates of end points A and B by ...
M = (A +B)/2
This can be solved for end point B, given end point A and midpoint M:
2M = A +B
2M -A = B
Then the generic equation for end point B given end point A and midpoint M is ...
B = 2M -A
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In each of the numbered problems, the midpoint and one end point are given. These can be matched with the variables in the above formula, so the solution is simply a matter of arithmetic. Beware all the minus signs.
10.
T = 2R -S
T = 2(-9, 4) -(2, -1) = (2(-9) -2, 2(4) +1)
T = (-20, 9)
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11.
R = 2S -T
R = 2(-4, -6) -(-7, -3) = (2(-4) +7, 2(-6) +3)
R = (-1, -9)
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12.
B = (A +C)/2 = ((-9, -4) +(-1, 6))/2 = (-9 -1, -4 +6)/2 = (-10, 2)/2
B = (-5, 1) . . . . the missing midpoint of AC
D = 2B -E . . . . the missing end point of BD
D = 2(-5, 1) -(-4, -3) = (2(-5) +4, 2(1) +3)) = (-10 +4, 2 +3)
D = (-6, 5)