A line is formed joining the two points P(−4, 1) and Q(11, −2). Straight line PQ intersects the x-axis at R. Find the :

(a) Since, R lies on the x-axis, its ordinate is zero.

Let the point R$(x, 0)$ divide the line segment joining P(−4, 1) and Q(11, −2) in the ratio $m : n$.

By section formula,

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

Substituting values, we get :

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

Hence, PR : RQ = 1 : 2.

(b) From part (a),

P(−4, 1), Q(11, −2) and $m : n$ = 1 : 2

By section formula,

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

Substituting values, we get :

$$\begin{aligned}\Rightarrow x = \frac{1(11) + 2(-4)}{1 + 2} \\[1em] \Rightarrow x = \frac{11 - 8}{3} \\[1em] \Rightarrow x = \frac{3}{3} \\[1em] \Rightarrow x = 1.\end{aligned}$$

R = $(x, 0)$ = (1, 0).

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

(c) By formula,

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

Substituting values, we get :

$$\begin{aligned}\Rightarrow m_{PQ} = \frac{-2 - 1}{11 - (-4)} \\[1em] \Rightarrow m_{PQ} = \frac{-3}{15} \\[1em] \Rightarrow m_{PQ} = -\frac{1}{5}\end{aligned}$$

We know that,

The product of the slopes of two perpendicular lines is −1.

∴ Slope of the line perpendicular to PQ = $\frac{-1}{-\frac{1}{5}}$ = 5

By point-slope form,

$y - y_1 = m(x - x_1)$

Equation of the line perpendicular to PQ and passing through R(1, 0) is :

$$\begin{aligned}\Rightarrow y - 0 = 5(x - 1) \\[1em] \Rightarrow y = 5x - 5 \\[1em] \Rightarrow 5x - y = 5.\end{aligned}$$

Hence, the equation of the line perpendicular to PQ at R is $\bm{5x − y = 5}$.