Show that x^2 + 4xy − 2y^2 + 6x − 12y − 15 = 0 represents a pair of straight lines and that these lines together with the pair of lines x^2 + 4xy − 2y^2 = 0 form a rhombus.


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Question Number 26411 by Tinkutara last updated on 25/Dec/17

$S h o w t h a t x^{2} + 4 x y - 2 y^{2} + 6 x - 12 y$ $- 15 = 0 r e p r e s e n t s a p a i r o f s t r a i g h t$ $l i n e s a n d t h a t t h e s e l i n e s t o g e t h e r$ $w i t h t h e p a i r o f l i n e s x^{2} + 4 x y - 2 y^{2}$ $= 0 f o r m a r h o m b u s .$
	Answered by ajfour last updated on 25/Dec/17

	$x^{2} + 4 x y - 2 y^{2} + 6 x - 12 y - 15 = 0$ $\| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix} \| = \| \begin{matrix} 1 & 2 & 3 \\ 2 & - 2 & - 6 \\ 3 & - 6 & - 15 \end{matrix} \|$ $= - 6 + 24 - 18 = 0$ $h e n c e p a i r o f s t r a i g h t l i n e s .$ $I f b o t h p a i r o f l i n e s h a v i n g$ $s a m e t w o s l o p e s, h a v e a c o m m o n$ $a n g u l a r b i s e c t o r, t h e n t h e y f o r m$ $a r h o m b u s . H e r e a s t h e o t h e r$ $p a i r p a s s e s t h r o u g h o r i g i n, w e$ $c a n p r o v e t h i s b y s h o w i n g o r i g i n$ $i s e q u i d i s t a n t f r o m t h e e a c h$ $l i n e o f t h e f i r s t p a i r .$ $\Rightarrow \frac{c_{1}}{\sqrt{1 + m_{1}^{2}}} = \frac{c_{2}}{\sqrt{1 + m_{2}^{2}}}$ $o r c_{1}^{2} (1 + m_{2}^{2}) = c_{2}^{2} (1 + m_{1}^{2}) . . (i)$ $(y - m_{1} x - c_{1}) (y - m_{2} x - c_{2}) = 0$ $c_{1} c_{2} = \frac{- 15}{- 2} = \frac{15}{2}; t h e y a r e r o o t s o f$ $z^{2} + 6 z + \frac{15}{2} = 0$ $c_{1}, c_{2} = \frac{- 6 \pm \sqrt{36 - 30}}{2} = - 3 \pm \frac{\sqrt{6}}{2}$ $m_{1} m_{2} = \frac{a}{b} = - \frac{1}{2},$ $m_{1} + m_{2} = - \frac{2 h}{b} = 2; \Rightarrow r o o t s o f$ $z^{2} - 2 z - \frac{1}{2} = 0$ $m_{1}, m_{2} = \frac{2 \pm \sqrt{4 + 2}}{2} = 1 \pm \frac{\sqrt{6}}{2}$ $A l s o m_{1} c_{2} + m_{2} c_{1} = \frac{2 g}{b} = \frac{6}{- 2} = - 3$ $(1 + \frac{\sqrt{6}}{2}) (- 3 - \frac{\sqrt{6}}{2}) + (1 - \frac{\sqrt{6}}{2}) (- 3 + \frac{\sqrt{6}}{2})$ $= - 3 - \frac{\sqrt{6}}{2} - \frac{3 \sqrt{6}}{2} - \frac{3}{2} - 3 + \frac{\sqrt{6}}{2} + \frac{3 \sqrt{6}}{2} - \frac{3}{2}$ $= - 9, s o$ $(1 + \frac{\sqrt{6}}{2}) (- 3 + \frac{\sqrt{6}}{2}) + (1 - \frac{\sqrt{6}}{2}) (- 3 - \frac{\sqrt{6}}{2})$ $= - \frac{3}{2} - \sqrt{6} - \frac{3}{2} + \sqrt{6} = - 3 . . . . (i i)$ $h e n c e i f m_{1} = 1 - \frac{\sqrt{6}}{2}, c_{2} = - 3 - \frac{\sqrt{6}}{2}$ $e q . o f a n g u l a r b i s e c t o r o f f i r s t$ $p a i r o f l i n e s a r e t h u s$ $\frac{y - m_{1} x - c_{1}}{\sqrt{1 + m_{1}^{2}}} = \pm \frac{y - m_{2} x - c_{2}}{\sqrt{1 + m_{2}^{2}}}$ $a s t o f o r m a r h o m b u s o n e$ $b i s e c t o r l i n e m u s t p a s s t h r o u g h$ $o r i g i n, s o$ $c_{1} \sqrt{1 + m_{2}^{2}} = c_{2} \sqrt{1 + m_{1}^{2}}$ $a s c_{1} c_{2} > 0$ $\Rightarrow c_{1}^{2} - c_{2}^{2} = m_{1}^{2} c_{2}^{2} - m_{2}^{2} c_{1}^{2}$ $(c_{1} + c_{2}) (c_{1} - c_{2}) = (m_{1} c_{2} + m_{2} c_{1}) \times$ $(m_{1} c_{2} - m_{2} c_{1})$ $l . h . s . = (- 6) (\sqrt{6}) = - 6 \sqrt{6}$ $r . h . s . = (- 3) (2 \sqrt{6}) = - 6 \sqrt{6} [s e e (i i)]$ $h e n c e t h e l i n e s f o r m a r h o m b u s .$
	Commented by Tinkutara last updated on 25/Dec/17
	Thank you Sir!
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