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Misc 3 - Lines which cut-off intercepts on axes whose sum

Misc 3 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 3 - Chapter 10 Class 11 Straight Lines - Part 3
Misc 3 - Chapter 10 Class 11 Straight Lines - Part 4

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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 –6, respectively. Equation of a line by intercept form is π‘₯/π‘Ž + 𝑦/𝑏 = 1 where a is x – intercept & b is y – intercept Given that sum of intercept is 1 i.e. a + b = 1 Product of intercept is βˆ’ 6 i.e. a Γ— b = βˆ’6 From (1) a + b = 1 a = 1 – b Putting value of a in (2) a Γ— b = βˆ’6 (1 – b) Γ— b = βˆ’6 b – b2 = βˆ’6 0 = b2 – b – 6 b2 – b – 6 = 0 b2 – 3b + 2b – 6 = 0 b(b – 3) + 2(b – 3) = 0 (b – 3) (b + 2) = 0 So, b = 3, & b = – 2 For b = 3 From (1) a + b = 1 a + 3 = 1 a = 1 – 3 a = βˆ’2 For b = –2 From (1) a + b = 1 a – 2 = 1 a = 2 + 1 a = 3 Hence a = βˆ’2, b = 3 & a = 3, b = βˆ’2 Now, finding equation of lines For a = βˆ’2, b = 3 π‘₯/π‘Ž + 𝑦/𝑏 = 1 π‘₯/( βˆ’2) + 𝑦/3 = 1 (3π‘₯ βˆ’ 2𝑦 )/( βˆ’6 ) = 1 3x βˆ’ 2y = βˆ’ 6 βˆ’3x + 2y = 6 For a = βˆ’2, b = 3 π‘₯/π‘Ž + 𝑦/𝑏 = 1 π‘₯/( βˆ’2) + 𝑦/3 = 1 (3π‘₯ βˆ’ 2𝑦 )/( βˆ’6 ) = 1 3x βˆ’ 2y = βˆ’ 6 βˆ’3x + 2y = 6 For a = βˆ’2, b = 3 π‘₯/π‘Ž + 𝑦/𝑏 = 1 π‘₯/( βˆ’2) + 𝑦/3 = 1 (3π‘₯ βˆ’ 2𝑦 )/( βˆ’6 ) = 1 3x βˆ’ 2y = βˆ’ 6 βˆ’3x + 2y = 6 Hence, equation of lines are βˆ’3x + 2y = 6 & 2x βˆ’ 3y = 6

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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.