Question

In a triangle ABC, D divides BC in the ratio 3 : 2 and E divides CA in the ratio 1 : 3. The lines AD and BE meet at H and CH meets AB in F. Find the ratio in which F divides AB.

• A. $AF:FB=2:1$
• B. $AF:FB=1:2$
• C. $AF:FB=2:3$
• D. $AF:FB=3:2$

Answer 1

The correct option is A $AF:FB=2:1$

Take $A$ as origin and the position vectors of $B$ and $C$ be $b$ and $c$.

Hence the position vectors of other points under given conditions are

$D=\frac{3c+2b}{5},E=\frac{1.0+3c}{4}=\frac{3}{4}c.$

Equations of $AD$ and $BE$ are

$AD$ is $r=t\frac{3c+2b}{5}$

$BE$ is $r=(1-s)b+s\cdot \frac{3}{4}c.$

They intersect at $H$.

Comparing coefficients of $b$ and $c$, we get

$\frac{2}{5}t=1-s,\frac{3}{5}t=\frac{3}{4}s.$

$\therefore s=\frac{4}{5}t.$

$\therefore \frac{2}{5}t+\frac{4}{5}t=1$

$\therefore t=\frac{5}{6},s=\frac{4}{6}$

Point $H$ is $\frac{3c+2b}{6}.$

Now $F$ is point of inersection of $AB$ and $CH$ whose equations are $r=tb$

and $r=(1-s)c+s\frac{3c+2b}{6}$

Comparing the coefficients,

$t=\frac{2s}{6}=\frac{s}{3}$ and $1-s+\frac{3}{6}s=0\Rightarrow s=2\Rightarrow t=\frac{1}{3}s=\frac{2}{3}$

$\therefore$ P.V. of $F=\frac{2}{3}b$ or $\overrightarrow{AF}=\frac{2}{3}b,\overrightarrow{FB}=b-\frac{2}{3}b=\frac{1}{3}b$

$\therefore AF:FB=2:1$