Question

The line segment joining the points P (3, 3) and Q (6, - 6) is trisected at the points A and B such that A is nearer to P. If A also lies on the line given by 2x + y + k = 0, find the value of k.

Answer 1

Here, given points are P (3, 3) and Q (6, - 6) 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 = 2 and n = 1,

 x₁ =  \(\frac{2\times3+1\times6}{2+1}\), y₁ = \(\frac{2\times3+1\times(-6)}{2+1}\)

∴ x₁ = 4 , y₁ = 0 

∴Coordinates of A is (4,0) 

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 × 4 + 0 + k = 0 

∴ k = - 8