Question

Equation of a line AB is $$x + 2y + 6 = 0$$. A perpendicular PQ is dropped on AB from the point $$\mathrm{P}(3, -2)$$ meeting AB at Q. Find the: 

1. equation of PQ. 
2. coordinates of the point Q.

Answer 1

i. The equation of line $$\mathrm{AB}$$ is $$x + 2y + 6 = 0$$:

$$2y = -x - 6$$

$$\implies y = -\frac{1}{2}x - 3$$ 

Slope of line $$\mathrm{AB}$$ is $$m_1 = -\frac{1}{2}$$.

Since $$\mathrm{PQ} \perp \mathrm{AB}$$, slope of $$\mathrm{PQ}$$ is:

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

$$= -\frac{1}{-\frac{1}{2}}$$

$$= 2$$

Equation of $$\mathrm{PQ}$$ passing through $$\mathrm{P}(3, -2)$$: 

$$y - (-2) = 2(x - 3)$$

$$y + 2 = 2x - 6$$

$$2x - y - 8 = 0$$

ii. Point $$\mathrm{Q}$$ is the intersection of lines $$\mathrm{AB}$$ and $$\mathrm{PQ}$$: 

$$x + 2y = -6 \quad ...(1)$$

$$2x - y = 8 \quad ...(2)$$

From (2), $$y = 2x - 8$$. 

Substitute in (1):

$$x + 2(2x - 8) = -6$$ 

$$x + 4x - 16 = -6$$

$$5x = 10$$

$$\implies x = 2$$

Now, $$y = 2(2) - 8$$

$$= 4 - 8$$

$$= -4$$ 

Therefore, the coordinates of point $$\mathrm{Q}$$ are $$(2, -4)$$.