The line segment joining the points A(3, -4) and B (-2, 1) is divided in the ratio 1 : 3 at point P in it. Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y = 4.

Given points, A(3, -4) and B(-2, 1)

By section formula, the co-ordinates of the point P which divides AB in the ratio 1: 3 is given by

$P = \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big) \\[1em] = \Big(\dfrac{1 \times -2 + 3 \times 3}{1 + 3}, \dfrac{1 \times 1 + 3 \times -4}{1 + 3}\Big) \\[1em] = \Big(\dfrac{-2 + 9}{4}, \dfrac{1 + -12}{4}\Big) \\[1em] = \Big(\dfrac{7}{4}, -\dfrac{11}{4}\Big)

Given line equation is,

5x – 3y = 4

3y = 5x - 4

y = $\dfrac{5}{3}x - \dfrac{4}{3}$

So, the slope of this line (m) = $\dfrac{5}{3}$

Let slope of perpendicular line be m₁.

Then,

⇒ m₁ × m = -1

⇒ m₁ $\times \dfrac{5}{3} = -1$

⇒ m₁ = $-\dfrac{3}{5}$.

Slope of the required line = $-\dfrac{3}{5}$.

By point-slope form,

Equation of line through P and slope = $-\dfrac{3}{5}$ is,

⇒ y – y₁ = m(x – x₁)

$\Rightarrow y - \Big(-\dfrac{11}{4}\Big) = -\dfrac{3}{5}\Big(x - \dfrac{7}{4}\Big) \\[1em] \Rightarrow \dfrac{4y + 11}{4} = -\dfrac{3}{5} \times \dfrac{4x - 7}{4} \\[1em] \Rightarrow 4y + 11 = -\dfrac{3}{5} \times (4x - 7) \\[1em] \Rightarrow 5(4y + 11) = -3(4x - 7) \\[1em] \Rightarrow 20y + 55 = -12x + 21 \\[1em] \Rightarrow 12x + 20y + 55 - 21 = 0 \\[1em] \Rightarrow 12x + 20y + 34 = 0 \\[1em] \Rightarrow 2(6x + 10y + 17) = 0 \\[1em] \Rightarrow 6x + 10y + 17 = 0.$

Hence, P = $\Big(\dfrac{7}{4}, -\dfrac{11}{4}\Big)$ the equation of required line is 6x + 10y + 17 = 0.