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 $Q$.
From: Mathematics, The Core Course for A-level, Bostock and Chandler. Chapter 4 Q15.
The answer is $(4,3.5)(7,5)$or$(0.4,-1.9)(2.2,9.8)$
I have no idea where to begin with this question.
I've tried letting $P=(x,12-x)$ and $Q=(x,\frac{3}{2}x-\frac{5}{2})$ and then used the distance formula to try and equate $|AQ|=2|AP|$ but didn't get anywhere.
How should I approach this problem?
We can set $P(p,\frac{3p-5}{2}),Q(q,12-q)$. (you have errors here.)
From $AQ=2AP$, we have the following two cases :
Case 1 : $P$ is the midpoint of the line segment $AQ$. $$p=\frac{1+q}{2}\quad \text{and}\quad \frac{3p-5}{2}=\frac{2+(12-q)}{2}$$ Solving the system gives $p=4, q=7$.
Case 2 : $A$ is on the line segment $PQ$ where $AQ:AP=2:1$. $$1=\frac{2\times p+1\times q}{1+2}\quad \text{and}\quad 2=\frac{1\times (12-q)+2\times \frac{3p-5}{2}}{1+2}$$ Solving the system gives $p=0.4,q=2.2$.