Question

find the value of k so that the area of triangle formed by the vertices (1,-1) ,(-4,2k) and (-k,-5) is 24

Answer 1

Answer:

Step-by-step explanation:

A (1,-1), B(-4,2k) and C (-k,-5)

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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