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

Answer 1

Answer: (C)

By family of lines required line is:
$(x+y-5)+\lambda (x-y+3)=0$
equation of other line is $\frac{x}{-2}+\frac{y}{-3}=1\Rightarrow 3x+2y+6=0$
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 $-4x=6y-20=0$
$2x-3y+10=0$