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


Transcript

Ex 4.2, 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 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 = 1 , y2 = 2 x3 = 3 , y3 = 6 & ∆ = 0 Putting values 0 = 1/2 |■8(𝑥&𝑦&1@1&2&1@3&6&1)| 0 = 1/2 (x|■8(2&1@6&1)|−y|■8(1&1@3&1)|+1|■8(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 = –2x + y 2x – y = 0 y = 2x Thus, the required equation of line joining is y = 2x

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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.