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 $2x - 3y + 7 = 0$