Answer :

**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(x_{1}, y_{1}) and B(x_{2},y_{2})

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 (x_{1} , y_{1}) and B(x_{2} , y_{2}) such that A is nearer to P.

By section formula,

For point A (x_{1}, y_{1}) of PQ, where m = 2 and n = 1,

∴ x_{1 =} 4, y_{1} = 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

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