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Question Number 28960 by Tinkutara last updated on 02/Feb/18

Commented by Tinkutara last updated on 02/Feb/18
Identify if the Statement 2 is true or false and give reasons.
	Answered by ajfour last updated on 02/Feb/18

	$I f t h r e e l i n e s b e e q u a l l y i n c l i n e d,$ $l e t t h e i r s l o p e s b e$ $m_{0}, m_{1} = \frac{m_{0} - \sqrt{3}}{1 + m_{0} \sqrt{3}}, m_{2} = \frac{m_{0} + \sqrt{3}}{1 - m_{0} \sqrt{3}}$ $Σ m_{1} = \frac{3 m_{0} (3 - m_{0}^{2})}{1 - 3 m_{0}^{2}} = 3 p$ $Σ m_{1} m_{2} = - 3$ $m_{1} m_{2} m_{3} = \frac{m_{0} (m_{0}^{2} - 3)}{1 - 3 m_{0}^{2}} = - p$ $e q . h a v i n g t h e s e t h r e e r o o t s$ $m^{3} - (Σ m_{1}) m^{2} + (Σ m_{1} m_{2}) m$ $- (Π m_{1}) = 0$ $m^{3} - 3 {p m}^{2} - 3 m + p = 0 . . . . (i)$ $g i v e n e q . i n q u e s t i o n i s$ $k (x^{3} - 3 {x y}^{2}) + y^{3} - 3 x^{2} y = 0$ $l e t \frac{y}{x} = m$ $\Rightarrow m^{3} - 3 {k m}^{2} - 3 m + k = 0 . . (i i)$ $(i) a n d (i i) a r e s a m e$ $e q u a t i o n s f o r p = k .$
	Commented by Tinkutara last updated on 03/Feb/18

	$m_{0}, m_{1} = \frac{m_{0} - \sqrt{3}}{1 + m_{0} \sqrt{3}}, m_{2} = \frac{m_{0} + \sqrt{3}}{1 - m_{0} \sqrt{3}}$ $W h y t h i s a s s u m p t i o n ?$
	Commented by ajfour last updated on 03/Feb/18

	$S e e Q . 29007$
	Commented by Tinkutara last updated on 03/Feb/18
	But isn't that means the three lines are equally inclined to one another? We are not given to which line they are inclined.
	Commented by ajfour last updated on 03/Feb/18

	$t o e a c h o t h e r, p a i r w i s e t h i s c a n$ $b e t h e o b v i o u s i n t e r p r e t a t i o n;$ $m e a n s a n g l e s o f i n c l i n a t i o n t o$ $e a c h o t h e r i s 120 °, e q u i v a l e n t l y$ $60 ° e v e n .$
	Commented by Tinkutara last updated on 04/Feb/18
	Thank you very much Sir! I got the answer.
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