The equation of the straight line perpendicular to the straight line $3 x+2 y=0$ and passing through the…
The equation of the straight line perpendicular to the straight line $3 x+2 y=0$ and passing through the point of intersection of the lines $x+3 y-1=0$ and $x-2 y+4=0$ is
$2 x-3 y+1=0$
$2 x-3 y+3=0$
$2 x-3 y+5=0$
$2 x-3 y+7=0$
Solution
The point of intersection of lines $x+3 y-1=0$ and $x-2 y+4=0$ is $(-2,1)$.
Let equation of line perpendicular to the given line is $2 x-3 y+\lambda=0$.
Since, it passes through $(-2,1)$.
$\begin{array}{lc}
\therefore & 2(-2)-3(1)+\lambda=0 \\
\Rightarrow & \lambda=7 \\
\therefore & \text { Required line is } 2 x-3 y+7=0
\end{array}$