P is a point on the x-axis which divides the line joining A(-6, 2) and B (9, -4). Find:

(a) Let the point P(x, 0) divide the line segment joining A(-6, 2) and B(9, -4) in the ratio m : n.

By section formula,

$y = \frac{my_2 + ny_1}{m + n}$

Substituting the values we get :

$$\begin{aligned}\Rightarrow 0 = \frac{m(-4) + n(2)}{m+n} \\[1em] \Rightarrow 0 = -4m + 2n \\[1em] \Rightarrow 4m = 2n \\[1em] \Rightarrow \frac{m}{n} = \frac{2}{4} \\[1em] \Rightarrow \frac{m}{n} = \frac{1}{2}.\end{aligned}$$

Hence, the ratio in which P divides the line segment AB is 1 : 2.

(b) From part (a),

A (-6, 2) and B (9, - 4)

m : n = 1 : 2

By section-formula,

$x = \frac{mx_2 + nx_1}{m + n}$

$$\begin{aligned}\Rightarrow x = \frac{1(9) + 2(-6)}{1+2} \\[1em] \Rightarrow x = \frac{9 - 12}{3} \\[1em] \Rightarrow x = \frac{-3}{3} \\[1em] \Rightarrow x = -1.\end{aligned}$$

P = (x, 0) = (-1, 0).

Hence, the coordinates of the point P are (-1, 0).

(c) By formula,

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

$$\begin{aligned}\Rightarrow m_{AB} = \frac{-4 - 2}{9 - (-6)} \\[1em] \Rightarrow m_{AB} = \frac{-6}{15} \\[1em] \Rightarrow m_{AB} = \frac{-2}{5}\end{aligned}$$

Line parallel to AB will have the same slope, so the slope of the new line is also m = $\frac{-2}{5}$.

By point-slope formula,

y - y₁ = m(x - x₁)

Line passing through (-3, -2) and parallel to AB is :

$$\begin{aligned}\Rightarrow y - (-2) = \frac{-2}{5}(x - (-3)) \\[1em] \Rightarrow y + 2 = \frac{-2}{5}(x + 3) \\[1em] \Rightarrow 5(y + 2) = -2(x + 3) \\[1em] \Rightarrow 5y + 10 = -2x - 6 \\[1em] \Rightarrow 5y + 10 + 2x + 6 = 0 \\[1em] \Rightarrow 2x + 5y + 16 = 0 \\[1em]\end{aligned}$$

Hence, the equation of the line is 2x + 5y + 16 = 0.