I am having difficulty with the algebra related to a coordinate geometry question. A line is drawn through the point $A=(1,2)$ to cut the line $2y = 3x-5$ in $P$ and the line $x+y = 12$ in $Q$.  If $AQ$ = $2AP$, find the coordinates of $P$ and $Q$.
Case 1: 

Since $Q(a,12-a)$ is on line $x+y=12$ then the midpoint $P({a+1\over 2},{14-a\over 2})$ of $AQ$ is on line $2y=3x-5$. So we have:

$$2{14-a\over 2} =3{a+1\over 2} -5$$

So we have $a= 7$ and thus $Q(7,5)$ and $P(4,{7\over 2})$.

Case 2: 

The midpoint $P'({a+1\over 2},{14-a\over 2})$ of $AQ$ is not on line $2y=3x-5$.
Since $A$ is the midpoint of $PP'$ we have $P({3-a\over 2},{a-6\over 2})$. Since $P$ is on $2y=3x-5$ we have:

$$2{a-6\over 2} =3{3-a\over 2}-5$$

so $a= {5\over 4}$...
Hint - 
As AQ=2AP hence AP=PQ. Therefore use midpoint formula
Now consider P($x_1 , y_1$) and Q($x_2 ,y_2$). 
Now slope AP= slope PQ = slope AQ . Now  P($x_1 , y_1$) satisfy equation 1 and and Q of another one. I hope you can solve forward. 
Remember it's just a hint. Next time write what y you have tried