Let A(-1,1), B(3,4) and C(2,0) be given three points.

Question

Let $A(-1,1),\ B(3,4)$ and $C(2,0)$ be given three points. A line $y=mx,\ m>0$ intersects lines $AC$ and $BC$ at points $P$ and $Q$ respectively. Let $A_1$ and $A_2$ be the areas of $	riangle ABC$ and $	riangle PQC$ respectively, such that $A_1=3A_2$, then the value of $m$ is equal to:

Accepted Answer

Line $AC$: through $(-1,1),(2,0)$: $x+3y=2$. Solving with $y=mx$: $P=\left(\dfrac2{3m+1},\dfrac{2m}{3m+1}\right)$.

Line $BC$: through $(3,4),(2,0)$: $y=4x-8$. Solving with $y=mx$: $Q=\left(\dfrac8{4-m},\dfrac{8m}{4-m}\right)$.

Using the area ratio condition $\dfrac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQC)}=3$ (via the determinant/area formula with common vertex $C$), and simplifying:
$$15m^2-11m-4=0 \implies m=1 \text{ or } -\frac4{15}$$

Since $m>0$: $m=1$.

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