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

Transcript

Ex 10.3, 12 Two lines passing through the point (2, 3) intersects each other at an angle of 60ยฐ. If slope of one line is 2, find equation of the other line. We know that Angle between 2 lines be tan ฮธ =|(๐‘š_2 โˆ’ ๐‘š_1)/(1 + ๐‘š_2 ๐‘š_1 )| Here m1 = Slope of one line = 2 ฮธ = 60ยฐ (given) We need to find m2 Putting the values tan 60ยฐ = |(๐‘š_2 โˆ’ 2)/(1 + 2 ร— ๐‘š_2 )| โˆš3 = |(๐‘š_2 โˆ’ 2)/(1 + 2๐‘š_2 )| |(๐‘š_2 โˆ’ 2)/(1 + 2๐‘š_2 )|= โˆš3 (๐‘š_2 โˆ’ 2)/(1 + 2๐‘š_2 ) = ยฑ โˆš3 So, (๐‘š_2 โˆ’ 2)/(1 + 2๐‘š_2 ) = โˆš3 and (๐‘š_2 โˆ’ 2)/(1 + 2๐‘š_2 ) = โ€“ โˆš3 Taking (๐’Ž_๐Ÿ โˆ’ ๐Ÿ)/(๐Ÿ + ๐Ÿ๐’Ž_๐Ÿ ) = โˆš๐Ÿ‘ m2 โˆ’ 2 = โˆš3(1 + 2m2) m2 โˆ’ 2 = โˆš3 + 2โˆš3m2 m2 โ€“ 2โˆš3m2 = โˆš3 + 2 m2 (1 โ€“ 2โˆš3) = 2 + โˆš3 m2 = (2 + โˆš3)/(1 โˆ’ 2โˆš3) Taking (๐’Ž_๐Ÿ โˆ’ ๐Ÿ)/(๐Ÿ + ๐Ÿ๐’Ž_๐Ÿ ) = โˆ’ โˆš๐Ÿ‘ m2 โˆ’ 2 = โˆ’ โˆš3(1 + 2m2) m2 โˆ’ 2 = โˆ’ โˆš3 โˆ’ 2โˆš3m2 m2 + 2โˆš3m2 = โˆ’โˆš3 + 2 m2 (1 + 2โˆš3) = 2 โˆ’ โˆš3 m2 = (2 โˆ’ โˆš3)/(1 + 2โˆš3) We know that equation of a line passing through (x1, y1) & having slope m is (y โˆ’ y1) = m(x โˆ’ x1) Equation of a line passing through (2, 3) & having slope (๐Ÿ + โˆš๐Ÿ‘)/(๐Ÿ โˆ’ ๐Ÿโˆš๐Ÿ‘) is (y โˆ’ 3) = ( (2 + โˆš3))/(1 โˆ’ 2โˆš3) (x โˆ’ 2) (1 โˆ’ 2โˆš3)(y โˆ’ 3) = (2 + โˆš3)(x โˆ’ 2) 1(y โˆ’ 3) โˆ’ 2โˆš3(y โˆ’ 3) = 2(x โ€“ 2) + โˆš3(x โˆ’ 2) y โ€“ 3 โˆ’ 2โˆš3y + 6โˆš3 = 2x โ€“ 4 + โˆš3x โˆ’ 2โˆš3 y โˆ’ 2โˆš3y โˆ’ 4x โˆ’ โˆš3x = โ€“ 6โˆš3 โ€“ 2โˆš3 โ€“ 4 + 3 y (1 โˆ’ 2โˆš3) โˆ’ x(โˆš3 + 2) = โ€“ 1 โ€“ 8โˆš3 1 + 8โˆš3 = x(โˆš3 + 2) + y (2โˆš3 โ€“ 1) (โˆš๐Ÿ‘ + 2)x + (2โˆš๐Ÿ‘ โ€“ 1)y = 1 + 8โˆš๐Ÿ‘ Equation of a line passing through (2, 3) & having slope (๐Ÿ โˆ’ โˆš๐Ÿ‘)/(๐Ÿ + ๐Ÿโˆš๐Ÿ‘) is (y โˆ’ 3) = (2 โˆ’ โˆš3)/(2โˆš3 + 1)(x โˆ’ 2) (2โˆš3 + 1) (y โˆ’ 3) = (2 โˆ’ โˆš3) (x โˆ’ 2) 2โˆš3 (y โˆ’ 3) + 1 (y โˆ’ 3) = 2(x โˆ’ 2) โˆ’ โˆš3(x โˆ’ 2) 2โˆš3 (y โˆ’ 3) + 1 (y โˆ’ 3) = 2(x โˆ’ 2) โˆ’ โˆš3(x โˆ’ 2) 2โˆš3y โˆ’ 6โˆš3 + y โˆ’ 3 = 2x โˆ’ 4 โˆ’ โˆš3x + 2โˆš3 2โˆš3y + y โˆ’ 6โˆš3 โˆ’ 3 = 2x โˆ’ โˆš3x โˆ’ 4 + 2โˆš3 (2โˆš3 + 1)y โˆ’ 6โˆš3 โˆ’ 3 = (2 โˆ’ โˆš3)x โˆ’ 4 + 2โˆš3 (2โˆš3 + 1)y = โˆ’ (โˆš3 โˆ’ 2)x โˆ’ 4 + 2โˆš3 + 6โˆš3 + 3 (2โˆš3 + 1)y + (โˆš3โˆ’2)x = 2โˆš3 + 6โˆš3 โˆ’ 4 + 3 (โˆš๐Ÿ‘โˆ’๐Ÿ)x + (2โˆš๐Ÿ‘ + 1)y = 8โˆš๐Ÿ‘ โˆ’ 1 Hence the equation of lines is (โˆš3 + 2)x + (2โˆš3 โ€“ 1)y = 1 + 8โˆš3 or (โˆš3 โˆ’ 2)x + (2โˆš3 + 1)y = 8โˆš3 โˆ’ 1

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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.