Question

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

Answer 1

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

Here, $x_1=3,y_1=-5,x_2=-4$ and $y_2=8$

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

The coordinates of P can be found using the section formula $\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}$

here, $m=k$ and $n=1$

Co-ordinates of P $=(\frac{K\times (-4)+1\times 3}{K+1},\frac{K\times 8+1\times (-5)}{K+1})=(\frac{-4K+3}{K+1},\frac{8K-5}{K+1})$

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

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

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

$\Rightarrow 4K-2=0$

$\Rightarrow 4K=2$

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

Thus, the required value of K is $\frac{1}{2}$