Question

If the family of straight lines $ax+by+c=0$. where $2a+3b=4c$ 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, $ax+by+c=0$ and $2a+3b=4c$
Putting $c=\frac{2a+3b}{4}$ in $ax+by+c=0$,
we get $ax+by+\frac{2a+3b}{4}=0$
$\Rightarrow 4ax+4by+2a+3b=0$
$\Rightarrow a(4x+2)+b(4y+3)=0$
$\Rightarrow (4x+2)+\frac{b}{a}(4y+3)=0$
This is of the form $L_1+\lambda L_2=0$
So, set of lines are
$4x+2=0$ and $4y+3=0$
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
$y+\frac{3}{4}=1\left(x+\frac{1}{2}\right)$
$\Rightarrow 4y=4x-1$
$\Rightarrow 4x-4y-1=0$
On solving $x+y+1=0$
and $4x-4y-1=0$,
we get $Q\left(-\frac{3}{8},-\frac{5}{8}\right)$