Question

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

Answer 1

We know that by section formula, the co-ordinates of the points which divide internally the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$ are

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

According to given conditions ,let's say coordinates of point $P$ are $(x,y)$

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

It is gien that point $P$ lies on the line $x-y+2=0$ ,

Thus,

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

$9k-1-8k-3=-2(k+1)$

$3k=2$

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

So , the value of $k$ is $\frac{2}{3}$