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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

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Ex 10.3, 10

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Ex 10.3, 18 Important

Last updated at Aug. 28, 2021 by Teachoo

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