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Ex 10.3, 3 (i) - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at Aug. 28, 2021 by Teachoo
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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 You are here
Ex 10.3, 3 (ii) Deleted for CBSE Board 2023 Exams
Ex 10.3, 3 (iii) Important Deleted for CBSE Board 2023 Exams
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. x 3 y + 8 = 0 x 3 y + 8 = 0 8 = x + 3 x + 3 = 8 Divide equation by (( 1)2 + ( 3)^2 ) = (1 + 3)= 4 = 2 ( + 3 )/2 = 8/2 ( )/2 + 3/2y = 4 x(( 1)/2) + y ( 3/2) = 4 Normal form of any line is x cos + y sin = p Comparing (1) & (2) p = 4 & cos = 1/2 & sin = 3/2 Now, finding = 180 60 = 120 So, the normal form of line is x cos 120 + y sin 120 = 4 Hence perpendicular distance from origin = p = 4 & angle between perpendicular & the + ve x-axis = = 120
