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.

Given: points P (3, 3) and Q (6, - 6).

Line 2x + y + k = 0

To find: The value of k.

Formula Used:

section formula:

If point P (x, y) divides the line segment A(x₁, y₁) and B(x₂,y₂)

Then the coordinates of P are:

Explanation:

Here, given points are P (3, 3) and Q (6, - 6) which is trisected at the points A (x₁ , y₁) and B(x₂ , y₂) such that A is nearer to P.

By section formula,

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

∴ x_{1 =} 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