Q, R and S are points on the line joint $P(a,x)$ and $Q(b,y)$ such that $PQ=RS=ST$, then $(\frac{5a+3b}{8},\frac{5x+3y}{8})$ is the midpoint of

Question

A)$PQ$

B)$QR$

C)$RS$

D)$ST$

Now there is one thing you may notice, and that is the question doesn’t tell $QR=PQ=RS=ST.$

So one may naturally assume that QR is not equal to those line segments. Here is the problem though, if it were given so, the answer matches perfectly ie.

$R$ would be the midpoint of $PT$, with the coordinates $(\frac{a+b}{2},\frac{x+y}{2})$ and $Q$ would be the midpoint of $PR$ with coordinates $(\frac{3a+b}{4},\frac{3x+y}{4})$

Then the midpoint of QR would be the point given in the question.

As I said, this is assuming that QR is indeed equal to the other line segments. There is no other way for the answer to arrive.

My teacher says it can be done without that specific piece of info (he won’t tell me how though, long story).

So one way is to prove that QR is equal to the rest of the line segments. I don’t think there is a way, but please let me know if there is.

If there isn’t, is there some sort of written proof as to why this question cannot be solved without the complete data so that I can prove him wrong?

Answer 1

I think there is a misprint in the question. The point $(b,y)$ is given to be $T$ (since that is what you use in your solution for the case when all the equalities are given). I am assuming that in the rest of the answer.

Take $P(0,0), Q(1,0), R(n,0), S(n+1,0) \text { and } T(n+2,0)$ (for some large $n$). Then the given conditions are met, but none of the answers match, nor is $QR$ equal to the rest of the line segments.