Question

Point A lies on the line segment PQ joining P(6, –6) and Q(–4, –1) in such a way that `(PA)/(PQ) = 2/5`. If the point A also lies on the line 3x + k(y + 1) = 0, find the value of k.

Answer 1

Let the coordinates of A be (x, y) Here `(PA)/(PQ) = 2/5`. 

So, PA + AQ = PQ

`⇒PA +AQ =(5PA)/2`   ...`[∵ PA = 2/5 PQ]`

` ⇒AQ = (5PA)/2 - PA`

`⇒ (AQ)/(PA) = 3/2 `

`⇒ (PA)/(AQ) = 2/3 `

Let (x, y) be the coordinates of A, which dives PQ in the ratio 2 : 3 internally Then using section formula, we get

` X = (2 xx (-4) +3 xx (6))/(2+3) = (-8+18)/5= 10/5 = 2`

`y = (2 xx (-1) + 3 xx(-6))/(2+3) = (-2-18)/5 = (-20)/5 = -4`

Now, the point (2, -4)  lies on the line 3x + k(y + 1) = 0, therefore

3 × 2 + k(-4 + 1) = 0

⇒ 3k = 6

`⇒ k =6/3 =2`

Hence, k = 2.