Q29 of 46 Page 1

Find the values of k so that the area of the triangle with vertices (1, - 1), ( - 4, 2k) and ( - k, - 5) is 24 sq. units.

As we know area of triangle formed by three points (x1, y1) , (x2, y2) and (x3, y3)



In this case,


Area = 24 sq. units


x1 = 1 x2 = - 4 x3 = - k


y1 = - 1 y2 = 2k y3 = - 5


putting this value in above eqn we get




48 = 2k2 + 3k + 21


2k2 + 3k - 27 = 0


2k2 + 9k - 6k - 27 = 0


k(2k + 9) - 3(2k + 9) = 0


(k - 3)(2k + 9) = 0


k = 3 or



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