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 $Q$. 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 $Q$.
I found the lengths of the lines $AQ$ and $AP$ in terms of $x$ and $y$ and used $AQ=2AP$ to get this equation $11x^2-40x-16=0$
Working: $AP=\sqrt{(x-1)^2+(\frac{3}{2}x-\frac{9}{2})^2}=\sqrt{\frac{1}{4}(13x^2-62x+85)}$
$AQ=\sqrt{(x-1)^2+(-x+10)^2}=\sqrt{2x^2-22x+101}$
$AQ=2AP\Rightarrow \sqrt{2x^2-22x+101}=2\sqrt{\frac{1}{4}(13x^2-62x+85)}$
$\Rightarrow 2x^2-22x+101=13x^2-62x+85\Rightarrow 11x^2-40x-16=0$
Using the quadratic formula I found $x=4,x=-\frac{4}{11}$
I then substituted these values into the two equations in the question to get the coordinates $(4,\frac{7}{2}),(-\frac{4}{11},-\frac{15}{11})$ or $(4,8),(-\frac{4}{11},\frac{136}{11})$
But the answers in the book are $(4,\frac{7}{2}),(7,5)$ or $(\frac{2}{5},-\frac{19}{10}),(\frac{11}{5},\frac{49}{5})$
I don't know where I went wrong in my method.
Edit: Found some errors in my method and calculation.
Using $x=4$ in $2y=3x-5$ gives $P(4,\frac{7}{2})$
Line with points $AP$ has equation $y-2=\frac{\frac{7}{2}-2}{3}(x-1)\Rightarrow y=\frac{1}{2}x+\frac{3}{2}$
$Q$ lies on the same line. Therefore, the intersection between $y=\frac{1}{2}x+\frac{3}{2}$ and $y+x=12$ is the point $Q$. Working: $\frac{1}{2}x+\frac{3}{2}=-x+12\Rightarrow x=7$ Therefore, $y=-7+12=5$. Point Q has coordinates $(7,5)$.
But $x=-\frac{4}{11}$ is not the other $x$ coordinate of $P$ according to the answers in the book. I have no idea where I went wrong at this point.
 A: 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 $$P(\frac 25,-\frac{19}{10}), Q(\frac{11}{5},\frac{49}{5})$$
A: Your initial equation setup is incorrect. Note that $P$ and $Q$ do not have the same $x$-coordinate! Because of this faulty assumption, anything you do after that will not lead to the right answer.
A good way to fix this is to give the coordinate different variable names. Let $P\left(a, \frac{3a-5}{2}\right)$ and $Q\left(b, 12-b \right)$ we have the distance formula
$$ \sqrt{(b-1)^2 + (10-b)^2} = 2\sqrt{(a-1)^2+\left(\frac{3a-5}{2}-2\right)^2} \tag{1} $$
Now use the fact that $A,\ P,\ Q$ are collinear to obtain a second equation
$$ \frac{10-b}{b-1} = \frac{\frac{3a-5}{2}-2}{a-1} \tag{2} $$
where the LHS is equal to the slope of $AQ$ and the RHS the slope of $AP$
Solve these two equations to obtain the $x$-coordinates of $P$ and $Q$
