Check sibling questions

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

Ex 4.3, 4 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.3, 4 - Chapter 4 Class 12 Determinants - Part 3

 

 

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Transcript

Ex 4.3, 4 (i) Find equation of line joining (1, 2) and (3, 6) using determinants Let L be the line joining the points (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 3 points collinear Area of triangle = 0 Area of triangle ≠ 0 Area of triangle ≠ 0 We know that Area of triangle is given by ∆ = 1/2 |■8(x1&y1&[email protected]&y2&[email protected]&y3&1)| Here, x1 = x , y1 = y x2 = 1 , y2 = 2 x3 = 3 , y3 = 6 & ∆ = 0 Putting values 0 = 1/2 |■8(𝑥&𝑦&[email protected]&2&[email protected]&6&1)| 0 = 1/2 (x|■8(2&[email protected]&1)|−y|■8(1&[email protected]&1)|+1|■8(1&[email protected]&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 = –2x + y 2x – y = 0 y = 2x Thus, the required equation of line joining is y = 2x 3 points collinear Area of triangle = 0 Area of triangle ≠ 0 We know that Area of triangle is given by ∆ = 1/2 |■8(x1&y1&[email protected]&y2&[email protected]&y3&1)| Here, x1 = x , y1 = y x2 = 3 , y2 = 1 x3 = 9 , y3 = 3 & ∆ = 0 Putting values 0 = 1/2 |■8(𝑥&𝑦&[email protected]&1&[email protected]&3&1)| 0 = 1/2 (x|■8(1&[email protected]&1)|−y|■8(3&[email protected]&1)|+1|■8(3&[email protected]&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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.