If $k x^{2} - 10 x y + 12 y^{2} + 5 x - 16 y - 3 = 0$ represents a pair of straight line, what is the value of $K$ ?

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Solution

$\begin{matrix} I f t h a t r e p r e s e n t s t w o s t r a i g h t l i n e, t h a t m e a n s t h e e q u a t i o n c a n b e r e d u c e d i n t o, y = m_{1} x + c_{1} a n d y = m_{2} x + c_{2} N o w, (y - m_{1} x - c_{1}) (y - m_{2} x - c_{2}) = 0 sin c e t h e c o e f f i c i e n t o f y^{2} i s 12. w e h a v e t o m u l t i p l y b o t h s i d e s b y 12. 12 (y - m_{1} x - c_{1}) (y - m_{2} x - c_{2}) = 0 w e k n o w m a n y t h i n g s a c t u a l l y t o f i n d o u t t h e c o n s tan t s . W e h a v e c_{1} c_{2} = - \frac{1}{4} c_{1} + c_{2} = \frac{4}{3} w e c a n f o r m a q u a d r a t i c e q u a t i o n w i t h c_{1}, c_{2} a s r o o t s : x^{2} - \frac{4}{3} x - \frac{1}{4} = 0 12 x^{2} - 16 x - 3 = 0 (6 x + 1) (2 x - 3) = 0 a s s u m e (c_{1}, c_{2}) = (\frac{3}{2}, - \frac{1}{6}) m_{1} + m_{2} = \frac{5}{6} - - (1) m_{1} c_{2} + m_{2} c_{1} = \frac{5}{12} - \frac{m_{1}}{6} + \frac{3 m_{2}}{2} = \frac{5}{12} - 2 m_{1} + 18 m_{2} = 5 - - - - (2) O n c o m b i n i n g b o t h e q u a t i o n s, w e g e t m_{1} = \frac{1}{2} a n d, m_{2} = \frac{1}{3} S o, t h e c o e f f i c i e n t o f x^{2} i s 12 m_{1} m_{2} = 2 H e n c e, t h e v a l u e o f k i s 2! \end{matrix}$

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