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

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

Let point P be (x, y) which divides AB in ratio of k:1, also point P lies on line x + y = 0

x = \( \frac{mx_2+nx_1}{m+n}\), y = \( \frac{my_2+ny_1}{m+n}\)

For point P on the line joined by the points A and B.

x = \(\frac{k\times(-4)+1\times3}{k+1}\), y = \(\frac{k\times8+1\times(-5)}{k+1}\)

Putting in given equation,

now (x,y) lies on the line x + y = 0

Therefore, the points will satisfy the equation.

Hence,

\(\frac {-4k+3}{k+1}+\frac{8k-8}{k+1}\) = 0

- 4 k + 3 + 8 k - 8 = 04k - 5 = 0 

∴ 4k = 5 

∴ k = \(\frac{5}2\)