Ex 4.3, 4 - Chapter 4 Class 12 Determinants - Part 4

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Ex 4.3, 4 - Chapter 4 Class 12 Determinants - Part 5

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Ex 4.3, 4 - Chapter 4 Class 12 Determinants - Part 6

  1. Chapter 4 Class 12 Determinants (Term 1)
  2. Serial order wise

Transcript

Ex 4.3, 4 (ii) Find equation of line joining (3, 1) and (9, 3) using determinants Let L be the line joining the (3, 1) & (9, 3) 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 |■8(x1&y1&1@x2&y2&1@x3&y3&1)| Here, x1 = x , y1 = y x2 = 3 , y2 = 1 x3 = 9 , y3 = 3 & ∆ = 0 Putting values 0 = 1/2 |■8(𝑥&𝑦&1@3&1&1@9&3&1)| 0 = 1/2 (x|■8(1&1@3&1)|−y|■8(3&1@9&1)|+1|■8(3&1@9&3)|) 0 = 1/2 (𝑥 "(1 – 3) – y (3 – 9) +1 (9 – 9) " ) 0 = 1/2 (x (–2) – y ( –6) + 1 (0)) 0 = 1/2 (–2x + 6y + 0 ) 2 × 0 = ( –2x + 6y) 0 = –2x + 6y 2x – 6y = 0 2(x – 3y) = 0 x – 3y = 0 Thus, the required equation of is x – 3y = 0

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.