Question

The equation of the straight line perpendicular to the straight line $3x,+2y=0$ and passing through, the point of intersection of the lines $x+3y-1=0$ and $x-2y+4=0$ is

• A. $2x-3y+1=0$
• B. $2x-3y+3=0$
• C. $2x-3y+5=0$
• D. $2x-3y+7=0$

Answer 1

Answer: (D)

The point of intersection of lines $x+3y-1=0$ and $X-2y+4=0$ is $(-2,1)$.
Let equation of line perpendicular to the given line is $2x-3y+\lambda =0$
Since, it passes through $(-2,1)$.
$\therefore 2(-2)=3(1)+\lambda =0$
$\Rightarrow \lambda =7$
$\therefore$ Required line is $2x-3y+7=0$