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 $P$ and $Q$.

Solution 1:

Here is a method you might prefer. We can write $$P(2s+1, 3s-1)$$ where $s$ is a parameter. Clearly the parametrization is not unique, and I have chosen this for convenience. Similarly we can write $$Q(t, 12-t)$$

Now $$\overrightarrow{AP}=\left(\begin{matrix}2s\\3s-3\end{matrix}\right)$$ and $$\overrightarrow{AQ}=\left(\begin{matrix}t-1\\10-t\end{matrix}\right)$$

Then $$\overrightarrow{AQ}=\pm2\overrightarrow{AP}\Rightarrow\left(\begin{matrix}t-1\\10-t\end{matrix}\right)=\pm2\left(\begin{matrix}2s\\3s-3\end{matrix}\right)$$

With the $+$ sign we get $s=1.5, t=7$ giving $$P(4, 3.5), Q(7, 5)$$ With the $-$ sign we get $s=-\frac{3}{10}, t=\frac{11}{5}$, giving $$P(\frac 25,-\frac{19}{10}), Q(\frac{11}{5},\frac{49}{5})$$