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Question
Using determinants, find the values of k, if the area of triangle with vertices (–2, 0), (0, 4) and (0, k) is 4 square units.
Solution
A(ΔABC ) = 4
1(0-0)-1(-2K)+1(-8-0)=±8
2K - 8 = ±8
2K = ±8 + 8
∴ K =`16/2` or K=`0/2 rArr` K =8 or K = 0
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