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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 7 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 three points 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 |■8(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 |■8(𝑥&𝑦&1@1&3&1@0&0&1)| 0 = 1/2 (𝑥|■8(3&1@0&1)|−𝑦|■8(1&1@0&1)|+1|■8(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 (1, 3) : x1 = 1, y1 = 3 B (0, 0) : x2 = 0, y2 = 0 D (k, 0) : x3 = k , y3 = 0 Area of triangle is ∆ = 1/2 |■8(x1&y1&1@x2&y2&1@x3&y3&1)| ± 3 = 1/2 |■8(1&3&1@0&0&1@k&0&1)| ± 3 = 1/2 (1|■8(0&1@0&1)|−3|■8(0&1@k&1)|+1|■8(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 6 = –3k 3k = 6 k = 6/(−3) = –2 –6 = – 3k –3k = –6 k = (−6)/(−3) = 2 Hence, the value of k is 2 or –2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.