Question

The line segment joining the points a (2,1) and b(5,-8) is trisected at the points p and q such that p is nearer to

a. if p also lies on the line given by 2x- y + k = 0, find the value of k.

Answer 1

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.
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Answer 2

point P = (x,y)
x = 4+5/3
= 3
y = 2+(-8)/3
= -2
now, 2x-y+k=0
6-2+k=0
= k = -4ans.