Last updated at Dec. 8, 2016 by Teachoo

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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 𝑦 – x + √3 𝑦 = 8 Divide equation by √(( − 1)2 + (√3)^2 ) = √(1 + 3)= √4 = 2 ( − 𝑥 + √3 𝑦)/2 = 8/2 ( − 𝑥)/2 + √3/2y = 4 x(( − 1)/2) + y (√3/2) = 4 Normal form of any line is x cos 𝜔 + y sin 𝜔 = p Comparing (1) & (2) p = 4 & cos 𝜔 = − 1/2 & sin 𝜔 = √3/2 Now, finding ω ∴ ω = 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 = 𝜔 = 120° 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 (0𝑥 + 𝑦)/1 = 2/1 0x + y = 2 The normal form of any line is x cos 𝜔 + y sin 𝜔 = p Comparing (1) & (2) p = 2 and cos 𝜔 = 0 & sin 𝜔 = 1 We know that cos 90° = 0 and sin 90° = 1 Thus, 𝜔 = 90° So, 𝜔 = 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 = 𝜔 = 90° 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 (𝑥 − 𝑦)/√2 = 4/√2 (𝑥 − 𝑦)/√2 = 4/√2 × √2/√2 (𝑥 − 𝑦)/√2 = (4√2)/2 𝑥/√2 – ( 𝑦)/√2 = 2√2 x(1/√2) + y(( − 1)/√2) = 2√2 Normal form of any line is x cos ω + y sin ω = p Comparing (1) & (2) p = 2√2 & cos ω = 1/√2 & sin ω = ( − 1)/√2 Finding ω ∴ ω = 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 = 𝜔 = 315°

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.