Question

$O(0,0),A(-3,-1)$ and $B(-1,-3)$ are the vertices of a $△OAB$. $P$ is a point on the perpendicular $AD$ drawn from $A$ on $OB$ such that $\frac{AP}{PD}=\frac{3}{4}$. Then the equation of the line $L$ parallel to $OB$ and passing through $P$, is

• A. $3x-y+3=0$
• B. $21x-7y+32=0$
• C. $15x-5y+32=0$
• D. $3x-y+35=0$

Answer 1

Answer: (B)

Slope of $OB=\frac{-3-0}{-1-0}=3$
$\therefore$ Slope of $AD=\frac{-1}{\text{ Slope of }OB}=\frac{-1}{3}$
Now, equation of $OB$ is given by
$y-0=3(x-0)$
$\Rightarrow y=3x\dots$..(i)
and equation of $AD$ is given by
$y+1=\frac{-1}{3}(x+3)$
$3y+3=-x-3$
$x+3y=-6\dots (ii)$
On solving Eq. (i) and Eq. (ii), we get
$D\left(\frac{-3}{5},\frac{-9}{5}\right)$
Now, $P$ divides $AD$ in the ratio $3:4$

$\therefore P(x,y)=\left(\frac{\frac{-9}{5}-12}{7},\frac{\frac{-27}{5}-4}{7}\right)=\left(\frac{-69}{35},\frac{-47}{35}\right)$
$\therefore$ Equation of line passing through $P\left(\frac{-69}{36},\frac{-47}{35}\right)$ and parallel to $OB$ is
$y+\frac{47}{35}=3\left(x+\frac{69}{35}\right)$
$\Rightarrow 35y+47=3(35x+69)$
$\Rightarrow 35y+47=105x+207$
$\Rightarrow 105x-35y+160=0$
$\Rightarrow 21x-7y+32=0$