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Ex 10.3
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 You are here
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
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Ex 10.3, 16 Important
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Ex 10.3, 18 Important
Last updated at Aug. 28, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
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. (ii) y 2 = 0 y 2 = 0 0x + y 2 = 0 0x + y = 2 Dividing both side by (02 + 12) = (0 + 1) = 1 (0 + )/1 = 2/1 0x + y = 2 The normal form of any line is x cos + y sin = p Comparing (1) & (2) p = 2 and cos = 0 & sin = 1 We know that cos 90 = 0 and sin 90 = 1 Thus, = 90 So, = 90 & p = 2 Thus, the normal form of line is x cos 90 + y sin 90 = 2 Hence, perpendicular distance from origin = p = 2 & angle between perpendicular & the + ve x-axis = = 90