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  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise

Transcript

Ex 10.1, 13 If three point (h, 0), (a, b) & (0, k) lie on a line, show that ๐‘Ž/โ„Ž + ๐‘/๐‘˜ = 1 . Let points be A (h, 0), B (a, b), C (0, k) Given that A, B & C lie on a line Hence the 3 points are collinear โˆด Slope of AB = Slope of BC We know that Slope of a line through the points (x1, y1), (x2, y2) is m = (๐‘ฆ_2 โˆ’ ๐‘ฆ_1)/(๐‘ฅ_2 โˆ’ ๐‘ฅ_1 ) Slope of line AB through the points A(h, 0), B(a, b) Here x1 = h & y1 = 0 x2 = a & y2 = b Putting values m = (๐‘ โˆ’ 0)/(๐‘Ž โˆ’ โ„Ž) m = ๐‘/(๐‘Ž โˆ’ โ„Ž) Slope of line BC through the points B(a, b) & C(0, k) Here x1 = a & y1 = b x2 = 0 & y2 = k Putting values m = (๐‘˜ โˆ’ ๐‘)/(0 โˆ’ ๐‘Ž) m = (๐‘˜ โˆ’ ๐‘)/(โˆ’๐‘Ž) Now, Slope of AB = Slope of BC ๐‘/(๐‘Ž โˆ’ โ„Ž) = (๐‘˜ โˆ’ ๐‘)/( โˆ’ ๐‘Ž) โ€“a(b) = (k โ€“ b) (a โ€“ h) โ€“ab = k(a โ€“ h) โ€“ b(a โ€“ h) โ€“ab = ka โ€“ kh โ€“ ab + bh โ€“ab + ab = ka โ€“ kh + bh 0 = ka + bh โ€“ kh ka + bh = kh Dividing both sides by kh ๐‘˜๐‘Ž/๐‘˜โ„Ž + ๐‘โ„Ž/๐‘˜โ„Ž = ๐‘˜โ„Ž/๐‘˜โ„Ž ๐‘Ž/โ„Ž + ๐‘/k = 1 Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.