Question

Equation of a straight line passing through the point of intersection of $x-y+1=0$ and $3x+y-5=0$ and perpendicular to one of them is:

• A. $x+y+3=0$
• B. $x+y-3=0$
• C. $x-3y-5=0$
• D. $x+3y+5=0$

Answer 1

The correct option is B $x+y-3=0$

Given lines are $3x+y-5=0\cdots \left(i\right)$

& $x-y+1=0\cdots \left(ii\right)$

Slope line $\left(i\right)=-3\&$ slope of line $\left(ii\right)=1$

Equation of any line through the point of intersection of

the given lines is $(3x+y-5)+k(x-y+1)=0$

Since this line is perpendicular to one of the given lines,

$\frac{3+k}{k-1}=-1$ or $\frac{1}{3}\Rightarrow k=-1$ or $-5$

$\therefore$ The required straight line is :

$x+y-3=0$ or $x-3y+5=0$