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Chapter 10 Class 11 Straight Lines (Term 1)

Serial order wise

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 β6, respectively. Equation of a line by intercept form is π₯/π + π¦/π = 1 where a is x β intercept & b is y β intercept Given that sum of intercept is 1 i.e. a + b = 1 Product of intercept is β 6 i.e. a Γ b = β6 From (1) a + b = 1 a = 1 β b Putting value of a in (2) a Γ b = β6 (1 β b) Γ b = β6 b β b2 = β6 0 = b2 β b β 6 b2 β b β 6 = 0 b2 β 3b + 2b β 6 = 0 b(b β 3) + 2(b β 3) = 0 (b β 3) (b + 2) = 0 So, b = 3, & b = β 2 For b = 3 From (1) a + b = 1 a + 3 = 1 a = 1 β 3 a = β2 For b = β2 From (1) a + b = 1 a β 2 = 1 a = 2 + 1 a = 3 Hence a = β2, b = 3 & a = 3, b = β2 Now, finding equation of lines For a = β2, b = 3 π₯/π + π¦/π = 1 π₯/( β2) + π¦/3 = 1 (3π₯ β 2π¦ )/( β6 ) = 1 3x β 2y = β 6 β3x + 2y = 6 For a = β2, b = 3 π₯/π + π¦/π = 1 π₯/( β2) + π¦/3 = 1 (3π₯ β 2π¦ )/( β6 ) = 1 3x β 2y = β 6 β3x + 2y = 6 For a = β2, b = 3 π₯/π + π¦/π = 1 π₯/( β2) + π¦/3 = 1 (3π₯ β 2π¦ )/( β6 ) = 1 3x β 2y = β 6 β3x + 2y = 6 Hence, equation of lines are β3x + 2y = 6 & 2x β 3y = 6