Question

The line segment joining the point A(2,1) and B(5,-8) is trisected by the points P and Q where P is nearer to A. If the point P also lies on the line 2x-y+k=0, find the value of K.

Answer 1

Answer:

Step-by-step explanation:

The given point are A(2, 1) and B(5, – 8).

Given, P and Q trisects the line segment AB.

∴ AP = PQ = QB

PB = PQ + QB = AP + AP = 2AP

AP : PB = AP : 2AP = 1 : 2

∴ P divides the line segment AB in the ration 1 : 2.

∴ Coordinates of P

P(3, – 2) lies on 2x – y + k = 0.

∴ 2 × 3 – (– 2) + k = 0

⇒ 6 + 2 + k = 0

⇒ 8 + k = 0

⇒ k = – 8

Thus, the value of k is – 8.

Answer 2

The line segment AB is trisected by points P and Q. So, AP: PQ = 1: 2

Using section formula, the coordinates of point P are (1x5+2x2/1+2, 1x(-8)+2x1/1+2) = (3, -2)

Now, it is given that P lies on the line 2x - y + k = 0.

Therefore, we have:

2x3 - (-2) + k = 0

6+2+k=0

k=-8