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Let A(1, 3), B(0, 0) and C(K, 0) be the vertices of triangle ABC of area 3 sq. units. Find k using determinant method.

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Let A(1, 3), B(0, 0) and C(K, 0) be the vertices of triangle ABC of area 3 sq. units. Find k using determinant method.

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Area of triangle 

A = 1/2[1(0 – 0) – 3(0 – K) + 1(0 – 0) 

A = [3K] Given , Area = 3 sq. units, 

Thus- 3k/2 = +3 3k/2 = -3 

⇒ 3k = 6, 3k = -6 

k = 2, k = -2

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