find the values of k so that area of triangle with vertices (1,-1) (-4,2k) (-k,-5) is 24 Sq units
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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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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
Mark as brainliest if it helped!
munnati22:
It was wrong i corrected it!
ya I hv also got this answer
Thank you
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