Check sibling questions

Β 

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


Transcript

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

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.