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Points P(-6,10), Q(0,6), and R are collinear. One of the points is the midpoint of the segment formed by the other two points. a. What are the possible coordinates of R? b. RQ=√208. Does this information affect the answer to part a? a. The possible coordinates of R are (, ) (Type an ordered pair. Use a comma to separate answers as needed.)​

Sagot :

Answer:

  a. (-12, 14), (6, 2), (-3, 8)

  b. yes; R(-12, 14)

Step-by-step explanation:

Points P(-6, 10), Q(0, -6), and R lie on a line, with one of them being the midpoint of the other two. You want to know the possible locations of R, and its location if RQ=√208.

Setup

Point M is the midpoint of AB when ...

  M = (A +B)/2

If M and A are given, then B is ...

  2M -A = B . . . . . . above equation solved for B

a. Possible locations of R

There are three choices for the location of R.

P is the midpoint

  R = 2P -Q = 2(-6, 10) -(0, 6) = (-12, 20-6)

  R = (-12, 14)

Q is the midpoint

  R = 2Q -P = 2(0, 6) -(-6, 10) = (6, 12 -10)

  R = (6, 2)

R is the midpoint

  R = (P +Q)/2 = ((-6, 10) +(0, 6))/2 = (-6, 16)/2

  R = (-3, 8)

The possible coordinates of R are (-12, 14), (6, 2), (-3, 8).

b. R for RQ=√208

The length of the given segment PQ is ...

  d = √((x2 -x1)² +(y2 -y1)²) . . . . . distance formula

  d = √((0 -(-6))² +(6 -10)²) = √(6² +(-4)²) = √(36 +16) = √52

This is half the length of RQ, so we must have P as the midpoint of RQ.

This distance information chooses one of the three points found in part (a), R(-12, 14).

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