Question

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.

Answer 1

Given the equation of line $2x+3y-5=0$

Let A(8,9) and B(2,1) be the points

Let the point of division be of the form K:1 = m:n

$Nowx=\frac{m{x}_2+n{x}_1}{m+n}$

$=\frac{K\times 2+1\times 8}{K+1}$

$y=\frac{m{y}_2+n{y}_1}{K+1}=\frac{K\times 1+1\times 9}{K+1}$

$\Rightarrow (x,y)=(\frac{2K+8}{K+1},\frac{K+9}{K+1})$

Substitute $(x,y)$ in the equation

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

$\Rightarrow 2(2K+8)+3(K+9)=5(K+1)$

$\Rightarrow 4K+16+3K+27=5K+5$

$\Rightarrow 2K+38=0\therefore K=-19$

$\therefore$ $K:1=-19:1=m:n$