Question

Find the equation of the straight line which passes through the point $P(2,6)$ and cuts the coordinate axes at the point $A$ and $B$ respectively so that $\frac{AP}{BP}=\frac{2}{3}.$

Answer 1

Straight line equation point $(2,6)$

$\frac{AP}{BP}=\frac{2}{3}$

by section formula

$p(2,6)=(\frac{0+3x}{5},\frac{2y+0}{5})$

$p(2,6)=(\frac{3x}{5},\frac{2y}{5})$

$\therefore \frac{3x}{5}=2$ & $\frac{2y}{5}=6$

$x=\frac{10}{3}$

$y=15$

Equation of line

$\therefore \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

$\therefore \frac{y-0}{15-0}=\frac{x-\frac{10}{3}}{0-\frac{10}{3}}$

$\therefore \frac{4}{15}=\frac{x-10/3}{-10/3}$

$\therefore \frac{4}{15}\times \frac{-10}{3}=x-10/3$

$\therefore \frac{-2y}{9}=\frac{3\times -10}{3}$

$\therefore -2y=3(3x+0)$

$\therefore -2y=9x-30$

$\therefore 9x+2y-30=0$.