# Ex 4.3, 3 (ii) - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Aug. 18, 2021 by

Last updated at Aug. 18, 2021 by

Transcript

Ex 4.3, 3 Find values of k if area of triangle is 4 square units and vertices are (ii) (-2, 0), (0, 4), (0, k) The area of triangle is given by ∆ = 1/2 |■8(x1&y1&1@x2&y2&1@x3&y3&1)| Here Area of triangle 4 square units Since area is always positive, ∆ can have both positive & negative signs ∴ Δ = ± 4 Putting x1 = −2, y1 = 0, x2 = 0, y2 = 4, x3 = 0 y3 = k ± 4 = 1/2 |■8(−2&0&1@0&4&1@0&k&1)| ± 4 = 1/2 (−2|■8(4&1@k&1)|−0|■8(0&1@0&1)|+1|■8(0&4@0&k)|) ± 4 = 1/2 (−2(4−𝑘)−0(0−0)+1(0−0)) ± 4 = 1/2 (−2(4−𝑘)) ± 4 = –1 (4 – k) ± 4 = –4 + k So, 4 = –4 + k and –4 = –4 + k Solving 4 = –4 + k 4 + 4 = k 8 = k k = 8 Solving –4 = –4 + k –4 + k = –4 k = –4 + 4 k = 0 So, the required value of k is k = 8 or k = 0

Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

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