Coordinate Geometry: PQRS Parallelogram

[Taylor_1989]

I would like someone to give this a quick check, I am really not sure if I am over thinking this question. I got the right ans, just would like a quick check of my method; big thanks in advance.

question: $P(-1,5), Q(8,10), R(7,5) & S(x,y)$ are the veritices of the parallelogram PQRS. Calculate the coordinate of S.

So all I did was workout the gradient of the line Q & R and applied that backwards from P, which gave the ans y=0 x=-2 which is correct, but I can't get rid of the feeling, that I have done something wrong but not sure what.

Is this the correct way to approach this type of question, seems a little basic to me which causes the concern.

 

[HallsofIvy]

Well, you are making the assumption that side QR is parallel to side PS- that is, that the vertices are given "cyclicly" which is the standard way and apparently happens to be true. but still an assumption. What you could have done was find the slopes of all the lines given by P, Q, and R to find the first pair of parallel sides.