I am having difficulty with the algebra related to a coordinate geometry question.

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 $Q$.

Answer 1

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

Answer 2

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 $a= {5\over 4}$...

Answer 3

Hint:

WLOG the equation of any straight line passing through $A(1,2)$ can be written as $$m=\dfrac{y-2}{x-1}$$

Find its intersections with the other two lines in terms of $m$

Now use $$AQ=2AP$$

Answer 4

A generic point of line $p:2y=3x-5$ has coordinates $P\left(t,\dfrac{3t-5}{2}\right)$

On line $q:x+y=12$ generic point is $Q(u,12-u)$

we have $$AP=\frac{1}{2} \sqrt{ \left|13 t^2-62 t+85\right|},\;AQ=\sqrt{2 u^2-22 u+101}$$

$AQ=2AP$ if $$\sqrt{2 u^2-22 u+101}=\sqrt{ \left|13 t^2-62 t+85\right|}$$ so it can be $$2 u^2-22 u+101=13 t^2-62 t+85\lor 2 u^2-22 u+101=-(13 t^2-62 t+85)$$ with the condition that $A,P,Q$ are collinear. Cross product of vectors $\vec {AP},\;\vec {PQ}$ is null.

that is $$-5 t u+23 t+11 u-29=0$$

$ \left\{ \begin{array}{l} 2 u^2-22 u+101=13 t^2-62 t+85 \\ -5 t u+23 t+11 u-29=0 \\ \end{array} \right. $

$t = 2/5, u = 11/5, t = 4, u =7$

So we have the points

$$P_1\left(\dfrac{2}{5},-\dfrac{19}{10}\right),\;Q_1\left(\frac{11}{5},\frac{49}{5}\right)$$

$$P_2\left(4,\frac{7}{2}\right),\;Q_2(7,5)$$

The second equations gives no real results

Hope this helps

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