Ex 4.3

Chapter 4 Class 12 Determinants
Serial order wise

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Ex 4.3, 3 Find values of k if area of triangle is 4 square units and vertices are (k, 0), (4, 0), (0, 2) The area of triangle is given by âˆ† = 1/2 |â– 8(x1&y1&[email protected]&y2&[email protected]&y3&1)| Here Area of triangle 4 square units Since area is always positive, âˆ† can have both positive & negative signs âˆ´ Î” = Â± 4. Putting x1 = k, y1 = 0, x2 = 4, y2 = 0, x3 = 0 y3 = 2 Â±4 = 1/2 |â– 8(k&0&[email protected]&0&[email protected]&2&1)| Â±4 = 1/2 |â– 8(k&0&[email protected]&0&[email protected]&2&1)| Â± 4 = 1/2 (k|â– 8(0&[email protected]&1)|âˆ’0|â– 8(4&[email protected]&1)|+1|â– 8(4&[email protected]&2)|) Â± 4 = 1/2 [ k (0 â€“ 2) â€“0 (4 â€“ 0) + 1 (8 â€“ 0) ] Â± 4 Ã— 2 = (k (â€“2) â€“ 0 + 1 (8)) Â± 8 = â€“2k + 8 So 8 = â€“2k + 8 or â€“ 8 = â€“2k + 8 Solving 8 = âˆ’2k + 8 8 â€“ 8 = â€“2k 0 = â€“2k k = 0/(âˆ’2) = 0 +Solving â€“8 = â€“2k + 8 â€“8 â€“ 8 = â€“2k â€“16 = â€“2k k = (âˆ’16)/(âˆ’2) = 8 So, the required value of k is k = 8 or k = 0