# Thread: Intersecting lines; distance ratios

1. ## Intersecting lines; distance ratios

I'm really stumped on this question, but I'm sure the answer is really simple!

"There exist two lines $\displaystyle 2y = 3x - 5$ and $\displaystyle x+y=12$. A line passing through a pt A $\displaystyle (1, 2)$ intersects these lines at the pts P and Q, respectively. The distance AQ is twice that of AP. What are the co-ordinates of P and Q?"

So far, I have said:

$\displaystyle AQ = 2 \cdot AP$.

So,

$\displaystyle AQ^2 = 4AP^2$.

If we denote the coordinates of P as $\displaystyle (a, b)$ and those of Q, $\displaystyle (c, d)$, then by Pythagoras' Theorem,

$\displaystyle AQ^2 = (c - 1)^2 + (d - 2)^2$

and $\displaystyle AP^2 = (a - 1)^2 + (b - 1)^2$.

We can therefore substitute these two equations into $\displaystyle AQ^2 = 4AP^2$ to give an equation in $\displaystyle a, b, c$ and $\displaystyle d$. Then, using the equations of the lines we can see that we can rewrite $\displaystyle b$ as

$\displaystyle b = \frac{3a - 5}{2}$

and $\displaystyle d$ as

$\displaystyle d = 12 - c$

which we can then use to derive an equation in terms of $\displaystyle a$ and $\displaystyle c$. But how to eliminate one of these variables eludes me! Your help is much appreciated.

2. Well you also know that $\displaystyle 2b=3a-5~\&~c+d=12$.

3. Originally Posted by Plato
Well you also know that $\displaystyle 2b=3a-5~\&~c+d=12$.
I thought I'd already mentioned using these results...?