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