Question

Point P divides the line segment joining the points A (-1,3) and B (9,8) such that $\frac{AP}{BP}=\frac{k}{1}$. If P lies on the line x - y + 2=0, find the value of k.

Answer 1

We know that by section formula, if point $(x,y)$ divides line joining two points in ratio $m:n$ $(x_1,y_1)$ and $(x_2,y_2)$

then $x=\frac{m{x}_2+n{x}_1}{m+n}$, $y=\frac{m{y}_2+n{y}_1}{m+n}$

$\therefore x=\frac{9k-1}{k+1}$, $y=\frac{8k+3}{k+1}$

This point lies on $x-y+2=0$

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

$9k-1-8k-3+2k+2=0$

$3k-2=0$

$k=\frac{2}{3}$