Question

$ln$ a triangle O$AB,E$ is the mid-point of $OB$ and $D$ is a point on $AB$ such that $AD$: $DB=2:1$. lf $OD$ and $AE$ interesect at $P$, then the ratio $\frac{OP}{PD}$ is

• A. $1:2$
• B. $3:2$
• C. $8:3$
• D. $4:3$

Answer 1

The correct option is B $3:2$

$\Delta OAB$ let $O=(0,0),A=(x_1,y_1),B=(x_2,y_2)$

Then, $E=(\frac{x_2}{2},\frac{y_2}{2})$,$D=(\frac{2x_2+x_1}{3},\frac{2y_2+y_1}{3})$

Equation of OD is: $y-0=\frac{2y_2+y_1}{2x_2+x_1}(x-0)$

$y=\frac{2y_2+y_1}{2x_2+x_1}x$........(1)

Equation of AE is: $(y-y_1)=\frac{y_2-2y_1}{x_2-2x_1}(x-x_1)$

From eq(1)

$\left(\frac{2y_2+y_1}{2x_2+x_1}\right)\left(x\right)-y_1=\frac{y_2-2y_1}{x_2-2x_1}(x-x_1)$

so $x=\frac{(2x_2+x_1)}{5}=\frac{3\left(\frac{2x_2+x_1}{3}\right)+2\left(0\right)}{(3+2)}$

so ratio $OP:PD=3:2$