Question

The line joining the points (2,1) and (5,-8) is trisected at the points P and Q. If point P lies on the line 2x - y + k = 0. Find the value of k.

Answer 1

Here, given points are P (2, 1) and Q (5, - 8) which is trisected at the points(say) A(x₁ , y₁) and B(x₂ , y₂)such that A is nearer to P.

By section formula,

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

For point A(x₁ , y₁) of PQ, where m = 1 and n = 2,

x₁ = \(\frac{1\times5+ 2\times2}{1+2}\), y₁ = \(\frac{1\times(-8)+2\times1}{1+2}\)

∴ x₁ = 3 , y₁ = -2 

∴Coordinates of A is (3,-2) 

It is given that point A lies on the line 2x - y + k = 0. 

So, 

substituting value of x and y as coordinates of A, 

2 × 3 – (-2) + k = 0 

∴ k = -8