Question

a point p divides the line segment joining the points a(3,-5) and b (-4,8) such that ap/pb=k/1. if p lies on the line x+y=0, then find the value of k.

Answer 1

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Answer 2

Answer:

$\text{The value of k is }\frac{1}{2}$

Step-by-step explanation:

Given 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, we have to find the value of k

By section formula,

when a point p(x,y) divides the line segment joining the points a(3,-5) and b(-4,8) in ratio m:n=k:1 then coordinates of p are

$p(x,y)=(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})=(\frac{-4k+3}{k+1},\frac{8k-5}{k+1})$

As the above point passing through the line x+y=0

hence, satisfy the above equation.

⇒ $\frac{-4k+3}{k+1}+\frac{8k-5}{k+1}=0$

⇒ $-4k+3+8k-5=0$

⇒ $4k-2=0$

⇒ $k=\frac{2}{4}=\frac{1}{2}$

$\text{The value of k is }\frac{1}{2}$