Find the ratio in which the line 2x+ 3y – 5 = 0

Let the line 2x + 3y – 5 = 0 divides the line segment joining the points A (8, – 9) and B (2,1) in the ratio λ: 1 at point P.

$\therefore$ Coordinates of $P=\left\{\frac{2 \lambda+8}{\lambda+1}, \frac{\lambda-9}{\lambda+1}\right\}$

$\left[\because\right.$ internal division $\left.=\left\{\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right\}\right]$

But $P$ lies on $2 x+3 y-5=0$

$\therefore$ $2\left(\frac{2 \lambda+8}{\lambda+1}\right)+3\left(\frac{\lambda-9}{\lambda+1}\right)-5=0$

$\Rightarrow \quad 2(2 \lambda+8)+3(\lambda-9)-5(\lambda+1)=0$

$\Rightarrow \quad 4 \lambda+16+3 \lambda-27-5 \lambda-5=0$

$\Rightarrow \quad 2 \lambda-16=0$

$\Rightarrow \quad \lambda=8 \Rightarrow \lambda: 1=8: 1$

So, the point $P$ divides the line in the ratio $8: 1$.

$\therefore$ Point of division $P=\left\{\frac{2(8)+8}{8+1}, \frac{8-9}{8+1}\right\}$

$\equiv\left(\frac{16+8}{9},-\frac{1}{9}\right)$

$\equiv\left(\frac{24}{9}, \frac{-1}{9}\right) \equiv\left(\frac{8}{3}, \frac{-1}{9}\right)$

Hence, the required point of division is $\left(\frac{8}{3}, \frac{-1}{9}\right)$.