Question

In a triangle $ABC,D$ and $E$ are points on $BC$ and $AC$ respectively, such that $BD=2DC,AE=3EC,$ Let $P$ be the point of intersection of $AD$ and $BE$. Then $\frac{BE}{PE}=$

• A. $2:3$
• B. $3:8$
• C. $8:3$
• D. $1:2$

Answer 1

The correct option is C $8:3$

Let p.v of $\overrightarrow{a}=\overrightarrow{a}$
p.v of $\overrightarrow{B}=\overrightarrow{b}$
p.v of $\overrightarrow{C}=\overrightarrow{c}$
p.v of $E=\frac{3\overrightarrow{c}}{4}$
p.v of $D=\frac{3\overrightarrow{c}}{4}$
$=\frac{2\overrightarrow{c}+\overrightarrow{b}}{3}$
$p$ lie on line $AD$ so p.v of $p=b\left(\frac{2\hat{c}+\hat{b}}{3}\right)$..............(1)
Let $\frac{BP}{PA}=K$
So p.v of $p=\frac{K\frac{3\hat{c}}{4}+\hat{b}}{K+1}$.............(2)
equate (1) and (2)
$K=\frac{8}{3}$