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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.
Solution
We have two points A (2, 1) and B (5,−8). There are two points P and Q which trisect the line segment joining A and B.
Now according to the section formula if any point P divides a line segment joining A (x1, y1) and B (x2, y2) in the ratio m: n internally then,
`"P"("x", "y") = (("n""x"_1 + "m""x"_2)/("m" +"n") ,("n""y"_1 + "m""y"_2)/("m"+"n"))`
The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2
Now we will use section formula to find the coordinates of unknown point A as,
`"P"("x"_1, "y"_1) = ((1(5) + 2 (2))/(1 + 2), (2(1)+1(-8))/(1+2))`
= (3, -2)
Therefore, co-ordinates of point P is(3,−2)
It is given that point P lies on the line whose equation is
2x - y + k = 0
Since, point P satisfies this equation.
2(3) - (-2) + k = 0
So,
k = -8
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