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 point $P$ and $Q$ respectively. Let $A_1$ and $A_2$ be the areas of $\Delta ABC$ and $\Delta PQC$ respectively, such that $A_1=3A_2,$ then the value of $m$ is equal to :

• A. $1$
• B. $3$
• C. $2$
• D. $\frac{4}{15}$

Answer 1

The correct option is A $1$


$A_1=\text{Area of}\Delta ABC=\frac{1}{2}\Vert \begin{array}{ccc}-1& 1& 1\\ 2& 0& 1\\ 3& 4& 1\end{array}\Vert$
$\Rightarrow A_1=\frac{13}{2}$
Equation of line $AC$ is $y-1=-\frac{1}{3}(x+1)$
Solving it with line $y=mx,$ we get $P(\frac{2}{3m+1},\frac{2m}{3m+1})$
Equation of line $BC$ is $y-0=4(x-2)$
Solving it with line $y=mx,$ we get $Q(\frac{-8}{m-4},\frac{-8m}{m-4})$

$A_2=\text{Area of}\Delta PQC=\frac{1}{2}\Vert \begin{array}{ccc}2& 0& 1\\ \frac{2}{3m+1}& \frac{2m}{3m+1}& 1\\ \frac{-8}{m-4}& \frac{-8m}{m-4}& 1\end{array}\Vert =\frac{A_1}{3}=\frac{13}{6}$
$\Rightarrow \frac{26m^2}{3m^2-11m-4}=\pm \frac{13}{6}$
$\Rightarrow 12m^2=\pm (3m^2-11m-4)$
Taking $+$ve sign,
$9m^2+11m+4=0$ (Rejected $\because m$ is imaginary)
Taking $-$ve sign,
$15m^2-11m-4=0$
$\Rightarrow m=1,-\frac{4}{15}$
$\Rightarrow m=1$ as $m>0$