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Class-12-science»Maths

Let L₁ be a straight line through the origin and L₂ be a straight line x + y = 1. If the intercepts made by the circle x² + y² -x + 3y =0 on L₁and L₂ are equal , then find the equation(s) which represents L₁ .

$G i v e n L_{1} b e a s t r a i g h t l i n e p a s \sin g t h r o u g h o r i g i n a n d L_{2} b e a s t r a i g h t l i n e x + y = 1 s o l e t e q u a t i o n o f l i n e L_{1} b e y = m x . A n d, t h e g i v e n e q u a t i o n o f c i r c l e b e x^{2} + y^{2} - x + 3 y = 0 S o, c e n t r e o f c i r c l e = (- g, - f) = (\frac{1}{2}, - \frac{3}{2}) N o w, int e r c e p t s m a d e b y t h e c i r c l e x^{2} + y^{2} - x + 3 y = 0 o n L_{1} a n d L_{2} a r e e q u a l . S o, l i n e s w i l l b e e q u i d i s \tan t f r o m t h e c e n t r e o f t h e c i r c l e . T h e r e f o r e, \frac{| \frac{1}{2} - \frac{3}{2} | - 1}{\sqrt{1^{2} + 1^{2}}} = \frac{| \frac{3 m}{2} + \frac{1}{2} |}{\sqrt{m^{2} + 1^{2}}} \Rightarrow \frac{2}{\sqrt{2}} = \frac{| 3 m + 1 |}{2 \sqrt{m^{2} + 1}} \Rightarrow \frac{4}{2} = \frac{{(3 m + 1)}^{2}}{4 (m^{2} + 1)} \Rightarrow 8 (m^{2} + 1) = {(3 m + 1)}^{2} \Rightarrow 8 m^{2} + 8 = 9 m^{2} + 1 + 6 m \Rightarrow m^{2} + 6 m - 7 = 0 \Rightarrow m^{2} + 7 m - m - 7 = 0 \Rightarrow m (m + 7) - 1 (m + 7) = 0 \Rightarrow (m + 7) (m - 1) = 0 \Rightarrow m = - 7 a n d m = 1 T h u s, e q u a t i o n o f l i n e L_{2} w i l l b e y = - 7 x o r y = x i . e ., y + 7 x = 0 o r y - x = 0$

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