Ex 4.2,1 (i) - Find Area of Triangle with vertices (1, 0), (6, 0), (4, - Ex 4.2

part 2 - Ex 4.2,1 (i) - Ex 4.2 - Serial order wise - Chapter 4 Class 12 Determinants

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Ex 4.2, 1 Find area of the triangle with vertices at the point given in each of the following: (1, 0), (6, 0), (4, 3) The area of triangle is given by āˆ† = šŸ/šŸ |ā– 8(š±šŸ&š²šŸ&šŸ@š±šŸ&š²šŸ&šŸ@š±šŸ‘&š²šŸ‘&šŸ)| Here, x1 = 1 , y1 = 0 x2 = 6 ,y2 = 0 x3 = 4 ,y3 = 3 āˆ† = šŸ/šŸ |ā– 8(šŸ&šŸŽ&šŸ@šŸ”&šŸŽ&šŸ@šŸ’&šŸ‘&šŸ)| = 1/2 (1|ā– 8(0&1@3&1)|āˆ’0|ā– 8(6&1@4&1)|+1|ā– 8(6&0@4&3)|) = 1/2 (1(0 – 3) – 0(6 – 4) + 1 (18 – 0)) = 1/2 (1(–3) + 0 + 1 (18) ) = 1/2 [–3 + 18 ] = šŸšŸ“/šŸ Thus, the required area of triangle is šŸšŸ“/šŸ square units

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