Question
Sun February 17, 2013 By: Anish Mishra
 

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

Expert Reply
Sun February 17, 2013
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
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