Question

Heya Frnds ...........☺

Try this ........

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⭐ Plz give proper solution.......

Answer 1

Let the given vertices be A(1,-1),B(-4,2k) and C(-k,-5).

Here x1 = 1, x2 = -4, x3 = -k, y1 = -1,y2 = 2k, y3 = -5.

Given Area = 24 sq.units.

We know that Area of triangle = 1/2(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))

                                       24       = 1/2((1(2k - (-5)) + -4(-5 - (-1)) + -k(-1 - 2k))

                                    24 * 2     =  2k + 5 + 20 - 4 + k + 2k^2

                                         48      =  2k^2  + 21 + 3k

                                       2k^2  + 3k - 27 = 0

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

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

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

                                      k = 3 ( or) k = -9/2


Hope this helps!