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Ex 10.3, 8 - Find equation of line perpendicular to x - 7y + 5 = 0

Ex 10.3, 8 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.3, 8 - Chapter 10 Class 11 Straight Lines - Part 3
Ex 10.3, 8 - Chapter 10 Class 11 Straight Lines - Part 4

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Ex 10.3, 8 Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3. Let equation of line AB be x – 7y + 5 = 0 Let line CD be perpendicular to line AB and having x-intercept 3 Since Line CD has x-intercept 3 So, line CD passes through the point (3, 0) We have to find equation of line CD, Finding slope of line AB x − 7y + 5 = 0 − 7y = −x − 5 − 7y = −(x + 5) 7y = (x + 5) y = 1/7 (x + 5) y = 𝑥/7 + 5/7 The above equation is of the form y = mx + c where m = slope of line Thus, slope of line AB = 1/7 Now, Given that line AB and line CD are perpendicular We know that, product of slope of perpendicular lines is –1 So, (Slope of line AB) × (Slope of line CD) = –1 1/7 × (Slope of line CD) = –1 Slope of line CD = 7 × –1 Slope of line CD = –7 Equation of a line passing through a point (x0, y0) & having slope m is (y – y0)= m(x – x0) Equation of line CD passing through point P(3, 0) & having slope −7 is (y − 0) = –7(x – 3) y = –7 (x – 3) y = –7x – (–7) 3 y = –7x + 21 y + 7x = 21 Thus, the required equation of line is y + 7x = 21

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.