Question

A point $P$ divides the line segment joining the points $A(3,-5)$ and $B(-4,8)$ such that $\frac{AP}{PB}=\frac{K}{1}.$ If $P$ lies on the line $x+y=0,$ then find the value of $K.$

Answer 1

The given points are $A(3,-5)$ and $B(-4,8).$

Let, $x_1=3,y_1=-5,x_2=-4,y_2=8$.

Since $\frac{AP}{PB}=\frac{K}{1}$, the point $P$ divides the line segment joining the points $A$ and $B$ in the ratio $K:1.$

Using section formula: $(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Here $m=K,n=1$

Coordinates of $P=(\frac{K(-4)+1\left(3\right)}{K+1},\frac{K\left(8\right)+1(-5)}{K+1})$

$\Rightarrow (\frac{-4K+3}{K+1},\frac{8K-5}{K+1})$

It is given that, $P$ lies on the line $x+y=0.$

Therefore,

$(\frac{-4K+3}{K+1}+\frac{8K-5}{K+1})=0$

$\Rightarrow \frac{-4K+3+8K-5}{K+1}=0$

$\Rightarrow 4K-2=0$

$\Rightarrow 4K=2$

$\Rightarrow K=\frac{1}{2}$

So, the value of $K$ is $\frac{1}{2}$.