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point ?

Sol.

Let the point divide the line segment joining A(-3, -1) and B(-8, -9) in the
ratio k :1.
Then, x 1 = -8, y 1 = -9, x 2 = -3, y 2 = -1
Using the section formula, we have,

-5k - 5 = -3k - 8
-5k + 3k = -8 + 5 = -3
-2k = -3
-2k = -3

k =
Hence, the required ratio is 3 : 2

Exa:10- Find the coordinates of the points of trisection of the line segment joining
the points A(6, -2) and B (-8, 10).

Sol. Let P and Q be the points of trisection so that AP = PQ = QB.
For P, m 1 : m 2 = AP : PB = 1 : 2, (x 1 , y 1 ) = (6, -2) and (x 2 , y 2 ) = (-8, 10)

x = ,

and y =

Therefore, point P =
For Q, m 1 : m 2 = AQ : QB = 2 : 1, (x 1 , y 1 ) = (6, -2) and (x 2 , y 2 ) = (-8, 10)

Therefore, Q =

Exa:12- Given the points A(-1, 6), B(5, 4), C(3, -2) and D(-3, 0), prove that ABCD
is a parallelogram using mid-point formula.

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