A line passing through the origin perpendicularly cuts the line 3x - 2y = 6 at point M. Find the co-ordinates of M.

Answer: Option B

Equation of line perpendicular to line 3x - 2y = 6 is 2x + 3y + p = 0

As, the line passes through origin (0, 0)
∴ 2 × 0 + 3 × 0 + p = 0
∴ p = 0
Now, the equation is 2x + 3y = 0
∴ 2x = -3y
x = $$\frac{-3}{2}$$ y
By putting this value in equation 3x - 2y = 6
$$\begin{array}{rl}& \Rightarrow 3\left(\frac{-3}{2}\right)y-2y=6\\ & \Rightarrow \frac{-9}{2}y-2y=6\\ & \Rightarrow \frac{-13y}{2}=6\\ & \Rightarrow y=-\frac{12}{13}\\ & \therefore x=\frac{-3}{2}\times \left(\frac{-12}{13}\right)\\ & x=\frac{18}{13}\\ & \therefore \text{Co - ordinate of point}\\ & \text{M}=\left[\frac{18}{13},-\frac{12}{13}\right]\end{array}$$