Question

If the family of straight lines $a x+b y+c=0$. where $2 a+3 b=4 c$ is concurrent at the point $P ( I , m )$, then the foot of the perpendicular drawn from $P$ to the line $x+y+1=0$ is (c) examsnet.com

• A. $\left(\frac{-3}{8}, \frac{-5}{8}\right)$
• B. $\left(\frac{-2}{5}, \frac{-3}{5}\right)$
• C. $(3.-4)$
• D. $(-5,4)$

Answer 1

Answer: (A)

We have, $a x+b y+c=0$ and $2 a+3 b=4 c$
Putting $c=\frac{2 a+3 b}{4}$ in $a x+b y+c=0$,
we get $a x+b y+\frac{2 a+3 b}{4}=0$
$\Rightarrow 4 a x+4 b y+2 a+3 b=0$
$\Rightarrow a(4 x+2)+b(4 y+3)=0$
$\Rightarrow (4 x+2)+\frac{b}{a}(4 y+3)=0$
This is of the form $L_{1}+\lambda L_{2}=0$
So, set of lines are
$4 x+2=0$ and $4 y+3=0$
These two lines intersect at $\left(\frac{-1}{2}, \frac{-3}{4}\right)$
$\therefore P(l, m)=\left(\frac{-1}{2}, \frac{-3}{4}\right)$

Equation of line passing through $\left(\frac{-1}{2}, \frac{-3}{4}\right)$ and
perpendicular to $x+y+1=0$ is
$y+\frac{3}{4}=1\left(x+\frac{1}{2}\right) $
$\Rightarrow 4 y=4 x-1 $
$\Rightarrow 4 x-4 y-1=0$
On solving $x+y+1=0$
and $4 x-4 y-1=0$,
we get $Q\left(-\frac{3}{8},-\frac{5}{8}\right)$