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. $\frac{4}{15}$
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$P\equiv \left(x_1,m{x}_1\right)$
$Q\equiv \left(x_2,m{x}_2\right)$
$A_1=\frac{1}{2}\left|\begin{array}{ccc}3& 4& 1\\ 2& 0& 1\\ -1& 1& 1\end{array}\right|-\frac{13}{2}$
$A_2=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& m{x}_1& 1\\ x_2& m{x}_2& 1\\ 2& 0& 1\end{array}\right|$
$A_2=\frac{1}{2}\left|2\left(m{x}_1-m{x}_2\right)\right|-m\left|x_1-x_2\right|$
$A_1=3A_2\Rightarrow \frac{13}{2}=3m\left|x_1-x_2\right|$
$\Rightarrow \left|x_1-x_2\right|-\frac{16}{6m}$
$AC:x+3y=2$
$BC:y=4x-8$
$P:x+3y=2\&y=mx\Rightarrow x_1=\frac{2}{1+3m}$
$Q:y=4x-8\&y=mx\Rightarrow x_2=\frac{8}{4-m}$
$\left|x_1-x_2\right|-\left|\frac{2}{1+3m}-\frac{8}{4-m}\right|$
$\left|\frac{-26m}{(1+3m)(4-m)}\right|=\frac{26m}{(3m+1)|m-4|}$
$=\frac{26m}{(3m+1)(4-m)}$
$\left|x_1-x_2\right|=\frac{13}{6m}$
$\frac{26m}{(3m+1)(4-m)}=\frac{13}{6m}$
$\Rightarrow 12m^2=-(3m+1)(m-4)$
$\Rightarrow 12m^2=-\left(3m^2-11m-4\right)$
$\Rightarrow 15m^2-11m-4=0$
$\Rightarrow 15m^2-15m+4m-4=0$
$\Rightarrow (15m+4)(m-1)=0$
$\Rightarrow m=1$