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Last updated at Feb. 4, 2020 by Teachoo

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Misc 3 Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and โ6, respectively. Equation of a line by intercept form is ๐ฅ/๐ + ๐ฆ/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 โ 6 i.e. a ร b = โ6 From (1) a + b = 1 a = 1 โ b Putting value of a in (2) a ร b = โ6 (1 โ b) ร b = โ6 b โ b2 = โ6 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 = โ2 For b = โ2 From (1) a + b = 1 a โ 2 = 1 a = 2 + 1 a = 3 Hence a = โ2, b = 3 & a = 3, b = โ2 Now, finding equation of lines For a = โ2, b = 3 ๐ฅ/๐ + ๐ฆ/๐ = 1 ๐ฅ/( โ2) + ๐ฆ/3 = 1 (3๐ฅ โ 2๐ฆ )/( โ6 ) = 1 3x โ 2y = โ 6 โ3x + 2y = 6 For a = โ2, b = 3 ๐ฅ/๐ + ๐ฆ/๐ = 1 ๐ฅ/( โ2) + ๐ฆ/3 = 1 (3๐ฅ โ 2๐ฆ )/( โ6 ) = 1 3x โ 2y = โ 6 โ3x + 2y = 6 For a = โ2, b = 3 ๐ฅ/๐ + ๐ฆ/๐ = 1 ๐ฅ/( โ2) + ๐ฆ/3 = 1 (3๐ฅ โ 2๐ฆ )/( โ6 ) = 1 3x โ 2y = โ 6 โ3x + 2y = 6 Hence, equation of lines are โ3x + 2y = 6 & 2x โ 3y = 6

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About the Author

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.