Finding other points on the line given the midpoint

[adillhoff]

Homework Statement

Given P(-5, 9) and Q(-8, -7), find a point R such that Q is the midpoint of PR

Homework Equations

$$d = \sqrt{(x+8)^2+(y+7)^2}$$

The Attempt at a Solution

Because Q is the midpoint of PR, I know that d(P, Q) = d(Q, R). I also know that d(P, R) = d(Q, R), which is $$\sqrt{265}$$.

The point of Q is $$(\frac{-13}{2}, 1)$$. I really don't know where to go from here. This problem is one of the chapter review questions and the book does not cover this specific type of problem in the examples nor did we cover it in our assignments.

 

[p21bass]

If Q is the midpoint of PR, per the problem statement, d(P,R) does NOT equal d(Q,R). I'm going to assume that that was a typo, and you meant d(P,Q) = d(Q,R).

I have no idea what you mean, by the point of Q is (-13/2, 1), as point Q is given as (-8,-7).

Since you have two points on a line, there's a couple of ways to go about this. You can determine the slope of the line, and use that to find a point on the other side of Q. Or, you could simply take the difference in the x- and y-coordinates of P&Q and add/subtract them from Q - giving you point R.

First thing you should probably do is graph points P and Q and construct a line, so that you have an idea of about where R should be.

 

[adillhoff]

Yes that was a typo. I also mean to say that the midpoint of PQ = (-13/2, 1). I completely over-analyzed this problem as I do with many problems. I took your advice and found point R(-11, -23) by taking the difference between the x- and y-coordinates of P & Q.

Thanks for your help!