Question

Let $ABC$ be a triangle and $P$ be a point inside $ABC$ such that $\overrightarrow{PA}+2\overrightarrow{PB}+3\overrightarrow{PC}=0$ The ratio of the area of $\Delta ABC$ to that of $\Delta APC$ is

• A. $2$
• B. $\frac{3}{2}$
• C. $\frac{5}{3}$
• D. $3$

Answer 1

Answer: (D)

Given,
Point $P$ inside the $\Delta ABC$ such that
$PA+2PB+3PC=0$

Let $PA=a-P$
$PB=b-P$
$PC=c-P$
$\therefore (a-p)+2(b-p)+3(c-P)=0$
$\Rightarrow a+2b+3c=3P$
$\Rightarrow p=\frac{a+2b+3c}{6}$
Area of $\Delta APC=\frac{1}{2}|a\times P+P\times c+c\times a|$
$=\frac{1}{2}\left|a\times \frac{\left(a+2b+3c\right)}{6}+\frac{\left(a+2b+3c\right)}{6}\times c+c\times a\right|$
$=\frac{1}{2\times 6}\left|2\left(a\times b\right)+3\left(a\times c\right)+\left(a\times c\right)+2\left(b\times c\right)+6c\times a\right|$
$=\frac{1}{6}\left|a\times b+b\times c+c\times a\right|$
$\therefore \frac{\text{Area of Δ ABC}}{\text{Area of Δ APC}}=\frac{\frac{1}{2}|a\times b+b\times c+c\times a|}{\frac{1}{6}|a\times b+b\times c+c\times a|}=3$
$\therefore$ Ratio $=3:1$