Last updated at Feb. 4, 2020 by Teachoo

Transcript

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

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Important You are here

Misc 4

Misc 5

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13

Misc 14

Misc 15 Important

Misc 16 Important

Misc 17

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22 Important

Misc 23

Misc 24 Important

Chapter 10 Class 11 Straight Lines (Term 1)

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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 10 years. He provides courses for Maths and Science at Teachoo.