Question

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

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

Answer 1

Answer: (B)

Slope of $O B=\frac{-3-0}{-1-0}=3$
$\therefore $ Slope of $A D=\frac{-1}{\text { Slope of } O B}=\frac{-1}{3}$
Now, equation of $O B$ is given by
$y-0=3(x-0) $
$\Rightarrow y=3 x \ldots$..(i)
and equation of $A D$ is given by
$y+1=\frac{-1}{3}(x+3) $
$3 y+3=-x-3$
$x+3 y=-6 \ldots ( ii )$
On solving Eq. (i) and Eq. (ii), we get
$D\left(\frac{-3}{5}, \frac{-9}{5}\right)$
Now, $P$ divides $A D$ 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 $O B$ is
$y+\frac{47}{35}=3\left(x+\frac{69}{35}\right)$
$\Rightarrow 35 y+47=3(35 x+69)$
$\Rightarrow 35 y+47=105 x+207$
$\Rightarrow 105 x-35 y+160=0$
$\Rightarrow 21 x-7 y+32=0$