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Question
The line segment joining the points P(3, 3) and Q(6, -6) is trisected at the points A and B such that Ais nearer to P. If A also lies on the line given by 2x + y + k = 0, find the value of k.
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Notes
We have two points P (3, 3) and Q (6,−6). There are two points A and B which trisect the line segment joining P and Q.
Let the co-ordinate of A be `A (x_1, y_1)`
Now according to the section formula if any point P divides a line segment joining `A(x_1, y_1)` and `B(x_2, y_2)` in the ratio m: n internally than,
`P(x,y) = ((nx_1 + mx_2)/(m + n), (ny_1 + my_2)/(m + n))`
The point A is the point of trisection of the line segment PQ. So, A divides PQ in the ratio 1: 2
Now we will use section formula to find the coordinates of unknown point A as,
`A(x,y ) = ((2(3) + 1(6))/(1 + 2)"," (2(3) + 1(-6))/(1 + 2))`
Therefore, co-ordinates of point A is(4, 0)
It is given that point A lies on the line whose equation is
2x + y + k = 0
So point A will satisfy this equation.
2(4) + 0 + k = 0
So,
k = -8
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Assertion (A): Mid-point of a line segment divides the line segment in the ratio 1 : 1
Reason (R): The ratio in which the point (−3, k) divides the line segment joining the points (− 5, 4) and (− 2, 3) is 1 : 2.
