Point A lies on the line segment PQ joining P(6, – 6) and Q(– 4, – 1) in such a way that . If the point A also lies on the line 3x + k(y + 1) = 0, find the value of k.

Let the point P(x,y) divides AB

Then


X = (m1x2 + m2x1)/ m1 + m2


= (2 × (– 4) + 3 × 6)/2 + 3


= (– 8 + 18) /5


= 10/5 = 2


Y = (m1y2 + m2y1)/ m1 + m2


= (2 × (– 1) + 3 x( – 6))/ 5


= (– 2 – 18)/5


= – 20 / 5 = – 4


If the point A also lies on the line 3x + k(y + 1) = 0


Then


3 × 2 + k(– 4 + 1) = 0


6 – 3k = 0


6 = 3k


k = 2


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