Question

find the values of K so that the area of triangle with vertices (1, - 1) (- 4,2 K) and (- K - 5) is 24 square units

Answer 1

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Answer 2

Answer:

K= -9/2 or 3

Step-by-step explanation:

ArΔABC = 1/2 [x1 (y2-y3) + x2 (y3-y1) + x3 (y1-y2) ]

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

        24 = 1/2 [2k+5 + 16 +(k+2k²)]

        24 = 1/2 (2k+5+16+k+2k²)

        24 = 1/2 (2k² +3k +21)

        48 = 2k² +3k + 21

        2k²+3k-27 = 0

       2k² + 9k - 6k - 27 = 0

 or   2k² - 6k + 9k - 27 =0

       2k(k-3) + 9(k-3) = 0

       (2k+9) (k-3) = 0

∴ k = -9/2 or 3

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