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Sidana [21]
3 years ago
15

Would anyone like to help me with my hw

Mathematics
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

7) x,y --> (x+5, y-2

8)(x,y) --> (x-4, y+1)

9)(x,y) --> (x, y+2)

10)(x,y) --> (x-5,y+4)

11)(x,y) --> (x-3, y+1)

12) (x,y) --> (x+4, y+1)

Step-by-step explanation:

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Find the missing endpoint if S is the midpoint RT
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Answer:

  10. T(-20, 9)

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  12. D(-6, 5)

Step-by-step 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.

__

<h3>generic solution</h3>

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

__

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.

<h3>10.</h3>

  T = 2R -S

  T = 2(-9, 4) -(2, -1) = (2(-9) -2, 2(4) +1)

  T = (-20, 9)

__

<h3>11.</h3>

  R = 2S -T

  R = 2(-4, -6) -(-7, -3) = (2(-4) +7, 2(-6) +3)

  R = (-1, -9)

__

<h3>12.</h3>

  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)

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