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Ex 10.2, 15 - Perpendicular from origin to a line at (-2, 9)

Ex 10.2, 15 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.2, 15 - Chapter 10 Class 11 Straight Lines - Part 3
Ex 10.2, 15 - Chapter 10 Class 11 Straight Lines - Part 4

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Ex 10.2, 15 The perpendicular from the origin to a line meets it at the point (– 2, 9), find the equation of the line. Let line OC be perpendicular to line AB at point C (–2, 9) i.e. OC perpendicular AB We know that If two lines are perpendicular then product of their slopes is equal to −1 ∴ Slope of OC × Slope of AB = −1 Finding slope of OC Slope of line passing through (x1, y1) & (x2, y2) = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) Slope of OC passing through (0, 0) and (–2, 9) Slope of OC = (9 − 0)/(−2 − 0) = 9/(−2) = (−9)/2 Now, Slope of AB × Slope of OC = −1 Slope of AB × (−9/2) = −1 Slope of AB = 2/9 Now, Point C(−2, 9) lie on the line AB Hence line AB passes through C(−2, 9) with slope 2/9 We know that equation of line passing through (x0, y0) & having slope m is (y – y0) = m (x – x0) Equation of line AB passing through (−2, 9)& having slope 2/9 (y – 9) = 2/9 (x – (−2)) y – 9 = 2/9 (x + 2) 9(y – 9) = 2(x + 2) 9y – 81 = 2x + 4 9y – 2x – 81 – 4 = 0 9y – 2x – 85 = 0 0 = 2x – 9y + 85 2x – 9y + 85 = 0 Which is the required equation of AB

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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.