Q38 of 149 Page 14

The line joining the points (2,1) and (5,-8) is trisected at the points P and Q. If point P lies on the line 2x - y + k = 0. Find the value of k.

Here, given points are P (2, 1) and Q (5, - 8) which is trisected at the points(say) A(x1 , y1) and B(x2 , y2)such that A is nearer to P.



By section formula,


x = , y =


For point A(x1 , y1) of PQ, where m = 1 and n = 2,


x1 = , y1 =


∴ x1 = 3 , y1 = -2


∴Coordinates of A is (3,-2)


It is given that point A lies on the line 2x - y + k = 0.


So, substituting value of x and y as coordinates of A,


2 × 3 – (-2) + k = 0


∴ k = -8


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