Example 18 - Chapter 4 Class 12 Determinants

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Example 18 Find the equation of the line joining A(1, 3) and B(0, 0) using determinants and find k if D(k, 0) is a point such that area of triangle ABD is 3 sq units. Equation of line Let L be the line joining the A(1, 3) & B(0, 0) Let (x, y) be the third point on line Since all the there point lie on the same line, they do not from a triangle Hence, Area of triangle = 0 Thus, ∆ = 0 We know that Area of triangle is given by ∆ = 1﷮2﷯ x1﷮y1﷮1﷮x2﷮y2﷮1﷮x3﷮y3﷮1﷯﷯ Here, ∆ = 0 x1 = x , y1 = y x2 = 1 , y2 = 3 x3 = 0 , y3 = 0 Putting values 0 = 1﷮2﷯ 𝑥﷮𝑦﷮1﷮1﷮3﷮1﷮0﷮0﷮1﷯﷯ 0 = 1﷮2﷯ x 3﷮1﷮0﷮1﷯﷯−y 1﷮1﷮0﷮1﷯﷯+1 1﷮3﷮0﷮0﷯﷯﷯ 0 = 1﷮2﷯ ( x (3 – 0) – y (1 – 0) +1 (0 – 0)) 0 = 1﷮2﷯ (x (3) – y (1) + 0) 0 = 1﷮2﷯ (3x – y) 2 × 0 = 3x – y 0 = 3x – y 3x – y = 0 y = 3x Thus, the equation of line joining A & B is y = 3x Also given a point D (k, 0) & Area of triangle ∆ ABD is 3 square unit Since, Area of triangle is always positive , ∆ can have both positive and negative sings ∴ ∆ = ± 3 We have A : x1 = 1, y1 = 3 B : x2 = 0, y2 = 0 B : x3 = k , y3 = 0 Area of triangle is ∆ = 1﷮2﷯ x1﷮y1﷮1﷮x2﷮y2﷮1﷮x3﷮y3﷮1﷯﷯ ± 3 = 1﷮2﷯ 1﷮3﷮1﷮0﷮0﷮1﷮k﷮0﷮1﷯﷯ ± 3 = 1﷮2﷯ 1 0﷮1﷮0﷮1﷯﷯−3 0﷮1﷮k﷮1﷯﷯+1 0﷮0﷮k﷮0﷯﷯﷯ ± 3 = 1﷮2﷯ ( 1 (0 – 0) – 3 (k – 0) +1 (0 – 0)) ±3 = 1﷮2﷯ (0 – 3 (k) + 0) ±3 = 1﷮2﷯ ( –3k) ± 6 = – 3k So, 6 = – 3k or – 6 = – 3k