Question

Equation of a straight line which passes through the point of intersection of the straight lines $x+y-5=$ 0 and $x-y+3=0$ and perpendicular to the straight line intersecting $x$-axis at the point $(-2,0)$ and the $y$-axis at the point $(0,-3)$, is

• A. $2 x-y+10=0$
• B. $x-3 y+10=0$
• C. $2 x-3 y+10=0$
• D. $2 x-3 y+12=0$

Answer 1

Answer: (C)

By family of lines required line is:
$(x+y-5)+\lambda(x-y+3)=0$
equation of other line is $\frac{x}{-2}+\frac{y}{-3}=1 \Rightarrow 3 x+2 y+6=0$
Both are $\perp$ hence $m_1 m_2=-1$
$\Rightarrow \left(\frac{\lambda+1}{1-\lambda}\right)\left(\frac{-3}{2}\right)=-1 $
$\Rightarrow 3 \lambda+3=-2+2 \lambda \Rightarrow \lambda=-5$
Hence required line $-4 x=6 y-20=0$
$2 x-3 y+10=0$