Coordinate Geometry Problem I have a question that I've started at school but had couldn't figure out what to do or where to start.
Sorry, I don't have the question written down, just the image. Help is much appreciated.

In the diagram to the right, point $P$ lies on the line $\displaystyle y = \frac{3}{2}x$ and is one vertex of square $PQRS$. Point $R$ has coordinates $(5,0)$.
a. Find the coordinates of point $P$.
b. If point $R$ lies on line $l$, and line $I$ divides quadrilateral $PQRS$ into two regions of equal area, find the equation for line $l$.

 A: Hint: The first big step to solving this problem is to figure out the $x$ and $y$ coordinate of $P$. Notice that there are two distinct constraints on $P$: it must lie on the line $y=\frac32x$, and it must be the corner of a square. For me, these problems are easier to think about if they are written entirely in equations. So my advice would be to figure out "what does it mean to be on the corner of that square?"
Keep in mind that you already know the coordinate of one corner, and you also know a few things about squares :)
For the second part, you should notice that it only takes two points to determine a line. If you choose a second point at random, this will be a rather difficult problem. So my advice would be to figure out "what is a good choice of a second point which will give more information?"
A: 
The approach Eric Stucky describes for the first question will run like this.  We are given that point $ \ R \ $ is located at $ \ (5 , 0) \ $ .  For lack of specific information at the moment, we'll use coordinates $ \ (X, Y) \ $ for point $ \ P \ $ , the diagonally-opposite vertex.  Since this point lies on the line $ \ y = \frac{3}{2} x \ , $ we can write its coordinates as $ \ (X \ , \ \frac{3}{2}X) \ . $  This means the height of the square $ \ PQRS \ $ is $ \ \frac{3}{2} X \ $ and its "base" $ \ QR \ $ (or $ \ PS \ $ ) is $ \ 5 - X \ . $  As it is a square, we have $ \ \frac{3}{2} X \ = \ 5 - X \ . $
Answering the second question is a bit more involved, as we need to bring in a few geometrical facts.  It will be helpful to identify the point of intersection between the line $  \ y = \frac{3}{2} x \  $ and the line $ \ l \ $ as point $ \ T \ $ .  We wish to locate $ \ T \ $ so that triangle $ \ ORT \ $ has half the area of quadrilateral $ \ ORSP \  . $  
In order to make such an assessment, we need to find the areas of these figures.  Quadrilateral $ \ ORSP \  $ is a trapezoid with parallel "bases" $ \ PS \ $ and $ \ OR \ $ and height $ \ RS \ , $ so its area is given by 
$$ A_{quad} \ = \ \frac{1}{2} \ \cdot \ \frac{3}{2}X \ \cdot \ [ \ 5 \ + \ (5 - X) \ ] \ . $$  
From the point $ \ T \ , $ we "drop an altitude" for the triangle $ \ ORT \ , $ the length of which we shall call $ \ h \ $ .  The "base" for this triangle is then $ \ OR \ $ which has length 5 , making the area of this triangle $ \ A_{tri} \ = \  \frac{1}{2}  \cdot  5  \cdot  h \ $ .  We thus require that the  altitude be
$$ h \ = \ \frac{2 \ A_{tri}}{5} \ = \ \frac{A_{quad}}{5} \ . $$
Since point $ \ T \ $ lies on the line $  \ y = \frac{3}{2} x \  , $ its coordinates are then $ \ (\frac{2}{3}h \ , \ h) \ . $  This makes the slope of line $ \ l \ $ , passing through points $ \ T \ $ and $ \ R \ , $
$$ m \ = \ \frac{0 \ - \ h }{5 \ - \ \frac{2}{3}h} \ \ . $$
We can at last write the equation of line $ \ l \ $ in "point-slope" form as $ \ y - 0 \ = \ m \ (x - 5 ) \ . $
[For the sake of indicating a result, the equation in "general form" is $ \ 12x \ + \ 17y \ = \ 60 \ . $ ]
