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But the given area of ABC = 12 units .

11 - 7k = 24 and 11 - 7k = -24
7k = 11 - 24 and - 7k = - 24 - 11
7k = -13 and - 7k = - 35

and k = 5

Hence, the values of k are , 5.
Exa8-If A, B and C are the points (-1, 5), (3, 1), (5, 7) respectively and D, E, F
are the middle points of BC, CA and AB respectively prove that area of ABC =
4 area of ( DEF)

D, E and F, then being the mid-points of sides BC, CA and AB respectively of ABC the

co-ordinates of D are ie (4, 4)

Co-ordinates of E are ie (2, 6)

Co-ordinates of F are ie (1, 3)
Area of a triangle is given by

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