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Ex 9.3, 6 Find the distance between parallel lines 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 We know that , distance between two parallel lines Ax + By + C1 = 0 & Ax + By + C2 = 0 is d = |𝐶_1− 𝐶_2 |/√(𝐴^2 + 𝐵^2 ) Equation of first line is 15x + 8y – 34 = 0 Above equation is of the form Ax + By + C1 = 0 where A = 15, B = 8 & C1 = − 34 Equation of second line is 15x + 8y + 31 = 0 Above equation is of the form Ax + By + C2 = 0 where A = 15 , B = 8 , C2 = 31 Distance between parallel lines 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 is d = |𝐶_1− 𝐶_2 |/√(𝐴^2 + 𝐵^2 ) Putting values d = |−34 − 31|/√(〖(15)〗^2 + (8)^2 ) d = |−34 − 31|/√(225 + 64) d = |−65|/√289 d = 65/√(17 × 17) d = 65/17 Thus, the required distance is 𝟔𝟓/𝟏𝟕 units

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.