The line segment joining the points A(2,1) and B(5,-8) is trisected at points P and Q such that P is nearer to A .If P lies on the line given by 2x-y+k =0 , find the value of 'k' . pls answer fast!
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2
are they are collinear to each other
then it's easy to find
then it's easy to find
1ak5hm335h:
yes they r
so its easy apply the the area folmuale
Ok
Answered by
13
Given that A and B is trisected at P and Q. That means AP: PQ will be 1:2.
We know that coordinates of points dividing line segment
= (mx2+nx1/m+n, my2 + ny1/m+n)
= (1 * 5 + 2 * 2/3, 1 * -8 + 2 * 1/3)
= (3,-2).
Given that P lies on 2x - y + k = 0.
= 2 * 3 - (-2) + k = 0
= 6 + 2 + k = 0
k = -8.
Hope this helps!
We know that coordinates of points dividing line segment
= (mx2+nx1/m+n, my2 + ny1/m+n)
= (1 * 5 + 2 * 2/3, 1 * -8 + 2 * 1/3)
= (3,-2).
Given that P lies on 2x - y + k = 0.
= 2 * 3 - (-2) + k = 0
= 6 + 2 + k = 0
k = -8.
Hope this helps!
but still they are collinear so you have to apply area formula but you have section fomula
No dude. We should apply this formula only
Thanks for the brainliest.
thank u for answering !!
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