#### Find the ratio in which the line 2x + 3y – 5 = 0 divides the line segment joining the points (8, –9) and (2, 1). Also find the coordinates of the point of division.

Solution

$Section\, formula\left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Let point p(x, y) divides the line segment joining the points A(8, –9) and B(2, 1) in ratio k : 1.
(x1, y1) = (8, -9)                       (x2, y2) = (2, 1)
m1 : m2 = k:1
using section formula we have
$P\left ( x,y \right )= \left [ \frac{2k+8}{k+1},\frac{k-9}{k+1} \right ]$
Given equation is 2x + 3y – 5 = 0       …(2)
Put values of x and y in eqn. (2)
$2\left [ \frac{2k+8}{k+1} \right ]+3\left [ \frac{k-9}{k+1} \right ]-5= 0$

2(2k + 3) + 3(k – 9) – 5(k + 1) = 0
4k + 16 + 3k – 27 – 5k – 5 = 0
2k – 16 = 0
k = 8
Hence, p divides the line in ration 8 : 1.
Put k = 5 in eqn. (1)
$\left ( x,y \right )= \left [ \frac{2\left ( 8 \right )}{8+1},\frac{8-9}{8+1} \right ]$
$= \left ( \frac{16+8}{9},\frac{-1}{9} \right )= \left ( \frac{24}{9} ,\frac{-1}{9}\right )= \left ( \frac{8}{3},\frac{-1}{9} \right )$
Required point is $P\left ( \frac{8}{3} ,\frac{-1}{9}\right )$