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

Transcript

Misc 3 Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and โ€“6, respectively. Equation of a line by intercept form is ๐‘ฅ/๐‘Ž + ๐‘ฆ/๐‘ = 1 where a is x โ€“ intercept & b is y โ€“ intercept Given that sum of intercept is 1 i.e. a + b = 1 Product of intercept is โˆ’ 6 i.e. a ร— b = โˆ’6 From (1) a + b = 1 a = 1 โ€“ b Putting value of a in (2) a ร— b = โˆ’6 (1 โ€“ b) ร— b = โˆ’6 b โ€“ b2 = โˆ’6 0 = b2 โ€“ b โ€“ 6 b2 โ€“ b โ€“ 6 = 0 b2 โ€“ 3b + 2b โ€“ 6 = 0 b(b โ€“ 3) + 2(b โ€“ 3) = 0 (b โ€“ 3) (b + 2) = 0 So, b = 3, & b = โ€“ 2 For b = 3 From (1) a + b = 1 a + 3 = 1 a = 1 โ€“ 3 a = โˆ’2 For b = โ€“2 From (1) a + b = 1 a โ€“ 2 = 1 a = 2 + 1 a = 3 Hence a = โˆ’2, b = 3 & a = 3, b = โˆ’2 Now, finding equation of lines For a = โˆ’2, b = 3 ๐‘ฅ/๐‘Ž + ๐‘ฆ/๐‘ = 1 ๐‘ฅ/( โˆ’2) + ๐‘ฆ/3 = 1 (3๐‘ฅ โˆ’ 2๐‘ฆ )/( โˆ’6 ) = 1 3x โˆ’ 2y = โˆ’ 6 โˆ’3x + 2y = 6 For a = โˆ’2, b = 3 ๐‘ฅ/๐‘Ž + ๐‘ฆ/๐‘ = 1 ๐‘ฅ/( โˆ’2) + ๐‘ฆ/3 = 1 (3๐‘ฅ โˆ’ 2๐‘ฆ )/( โˆ’6 ) = 1 3x โˆ’ 2y = โˆ’ 6 โˆ’3x + 2y = 6 For a = โˆ’2, b = 3 ๐‘ฅ/๐‘Ž + ๐‘ฆ/๐‘ = 1 ๐‘ฅ/( โˆ’2) + ๐‘ฆ/3 = 1 (3๐‘ฅ โˆ’ 2๐‘ฆ )/( โˆ’6 ) = 1 3x โˆ’ 2y = โˆ’ 6 โˆ’3x + 2y = 6 Hence, equation of lines are โˆ’3x + 2y = 6 & 2x โˆ’ 3y = 6

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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.