# Ex 4.3, 4 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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 (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 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 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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.