find thr value of k,is the area of a triangle is 4 sq units and vertices are (k,o),(4,0),(0,2).
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4 = 1/2 [x1 (y2+y3) - x2 (y3+y1) - x3 (y1+y2)]
4 = 1/2 [k (0+2) - 4 (2+0) - 0 (0+0)]
4 = 1/2 [2K - 8 - 0]
4 = 1/2 x 2K - 8
4 + 8 = 1/2 x K
12 x 2 /2 = k
12 = K
4 = 1/2 [k (0+2) - 4 (2+0) - 0 (0+0)]
4 = 1/2 [2K - 8 - 0]
4 = 1/2 x 2K - 8
4 + 8 = 1/2 x K
12 x 2 /2 = k
12 = K
rohit656311:
wrong answer
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