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Ex 4.3, 4 - (i) Find equation of line joining (1, 2), (3, 6) - Equation of line using determinant

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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Ex 4.3, 4 (i) Find equation of line joining (1, 2) and (3, 6) using determinants Let L be the line joining the (1, 2) & (3, 6) 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, x1 = x , y1 = y x2 = 1 , y2 = 2 x3 = 3 , y3 = 6 & ∆ = 0 Putting values 0 = 1﷮2﷯ 𝑥﷮𝑦﷮1﷮1﷮2﷮1﷮3﷮6﷮1﷯﷯ 0 = 1﷮2﷯ x 2﷮1﷮6﷮1﷯﷯−y 1﷮1﷮3﷮1﷯﷯+1 1﷮2﷮3﷮6﷯﷯﷯ 0 = 1﷮2﷯ (x(2 – 6) – y (1 – 3) +1 (6 – 6)) 0 = 1﷮2﷯ (x ( – 4) – y ( – 2) + 1 (0) 0 = 1﷮2﷯ ( – 4x + 2y + 0 ) 0 = 1﷮2﷯ × 2 ( – 2x + y) 0 = – 2x + y 2x – y = 0 y = 2x Thus, the required equation of line joining is y = 2x 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﷯ 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﷯ 𝑥﷮𝑦﷮1﷮3﷮1﷮1﷮9﷮3﷮1﷯﷯ 0 = 1﷮2﷯ x 1﷮1﷮3﷮1﷯﷯−y 3﷮1﷮9﷮1﷯﷯+1 3﷮1﷮9﷮3﷯﷯﷯ 0 = 1﷮2﷯ (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 is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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