Geometry, finding the possible values of 'a' $P (a,4)$, $Q (2,3)$, $R (3,-1)$ and $S (-2,4)$ are four points.
If $|PQ| = |RS|$, find the possible values of a
I know this is a pretty basic problem but I'm having a lot of trouble with it, here is my answer:
I found $|RS|$ to be $\sqrt{50}$  so |PQ| = sqrt-2a^2 + 1 = sqrt50
and from there I got $\sqrt{8}$ which is incorrect because that is only one value. 
The true answer should be; $-5$ or $9$ (according to textbook)
What is the proper method that I should use to come to the correct solution?

There's two things that have gone wrong here.
The first is that you have incorrectly calculated the horizontal distance (I'll call it $d$) between P and Q.  It should be
$$\sqrt{50} = \sqrt{d^2+(4-3)^2}$$
$$50 = d^2+1^2$$
$$d^2=49$$
$$d=7$$
The second is, once you've gotten a distance value, you can then use it to find the possible values of $a$: it could be $2 + 7 = 9$ or $2 - 7 = -5$, as both are 7 units away from Q horizontally.
