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Ex 10.3, 3 (iii) - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at March 30, 2023 by Teachoo
Ex 10.3
Ex 10.3, 1 (i)
Ex 10.3, 1 (ii) Important
Ex 10.3, 1 (iii)
Ex 10.3, 2 (i)
Ex 10.3, 2 (ii)
Ex 10.3, 2 (iii) Important
Ex 10.3, 3 (i) Deleted for CBSE Board 2023 Exams
Ex 10.3, 3 (ii) Deleted for CBSE Board 2023 Exams
Ex 10.3, 3 (iii) Important Deleted for CBSE Board 2023 Exams You are here
Ex 10.3, 4
Ex 10.3, 5 Important
Ex 10.3, 6 (i) Important
Ex 10.3, 6 (ii)
Ex 10.3, 7
Ex 10.3, 8 Important
Ex 10.3, 9 Important
Ex 10.3, 10
Ex 10.3, 11
Ex 10.3, 12 Important
Ex 10.3, 13
Ex 10.3, 14 Important
Ex 10.3, 15
Ex 10.3, 16 Important
Ex 10.3, 17 Important
Ex 10.3, 18 Important
Transcript
Ex10.3, 3 Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis. (iii) x y = 4 x y = 4 Dividing both side by (12 + ( 1)2)= (1 + 1)= 2 ( )/ 2 = 4/ 2 ( )/ 2 = 4/ 2 2/ 2 ( )/ 2 = (4 2)/2 / 2 ( )/ 2 = 2 2 x(1/ 2) + y(( 1)/ 2) = 2 2 Normal form of any line is x cos + y sin = p Comparing (1) & (2) p = 2 2 & cos = 1/ 2 & sin = ( 1)/ 2 Finding = 360 45 = 315 Thus, the normal form of line is x cos 315 + y sin 315 = 2 2 Hence, perpendicular distance from origin = p = 2 2 & angle between perpendicular & the + ve x-axis = = 315
