the line segment joining the points A ( 2 , 1 )  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 2 x - y + k =0 , find the value of k. 

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 2xy + 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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As, AP=QP=QB 

so, AP=1 & PB=2 so, m1=1 & m2=2

P=> x= (m1x2+m2x1) / m1+my=(m1y2+m2y1)/m1+m2

  = (1*5 + 2*2)/3  =(1*-8)+(2*1)/3

  = 9/3  = -6/3

  =3  = -2

So, X=3 and Y=-2

By putting the values of x n y in the eq.

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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