Question

A line through origin meets the line $$2x = 3y + 13$$ at right angles at point Q. Find the co-ordinates of Q. 

[Hint. Find the co-ordinates of the foot of the perpendicular from $$P(0, 0)$$ to $$2x = 3y + 13$$.]

Answer 1

The given line is $$2x - 3y = 13$$: 

$$3y = 2x - 13$$

$$\implies y = \frac{2}{3}x - \frac{13}{3}$$

Slope $$m_1 = \frac{2}{3}$$

The perpendicular line passing through the origin $$(0,0)$$ has slope:

$$m = -\frac{1}{m_1}$$

$$= -\frac{3}{2}$$

Equation of the perpendicular line through origin:

$$y - 0 = -\frac{3}{2}(x - 0)$$

$$2y = -3x$$

$$3x + 2y = 0 \quad ...(1)$$

The point $$\mathrm{Q}$$ is the intersection of $$2x - 3y = 13$$ and $$3x + 2y = 0$$. 

From (1), $$y = -\frac{3}{2}x$$. 

Substitute into the first line:

$$2x - 3\left(-\frac{3}{2}x\right) = 13$$

$$2x + \frac{9}{2}x = 13$$

$$\frac{13}{2}x = 13$$

$$x = 2$$

Now find $$y$$: 

$$y = -\frac{3}{2}(2)$$

$$= -3$$

Thus, the coordinates of $$\mathrm{Q}$$ are $$(2, -3)$$.