If one line of the equation : ax^3 +bx^2 y+cxy^2 +dy^3 =0 bisects the angle between the the other two then prove (3a+c)^2 (bc+2cd−3ad)= (b+3d)^2 (bc+2ab−3ad) .


Question and Answers Forum

All Questions Topic List
Coordinate Geometry Questions
Previous in All Question Next in All Question
Previous in Coordinate Geometry Next in Coordinate Geometry
Question Number 28303 by ajfour last updated on 24/Jan/18

$I f o n e l i n e o f t h e e q u a t i o n :$ ${a x}^{3} + {b x}^{2} y + {c x y}^{2} + {d y}^{3} = 0$ $b i s e c t s t h e a n g l e b e t w e e n t h e$ $t h e o t h e r t w o t h e n p r o v e$ ${(3 a + c)}^{2} (b c + 2 c d - 3 a d) =$ ${(b + 3 d)}^{2} (b c + 2 a b - 3 a d) .$
	Answered by ajfour last updated on 24/Jan/18

	$l e t t h e l i n e s b e$ $y = m_{1} x, y = m_{0} x, a n d y = m_{2} x$ $θ_{0} = (\frac{θ_{1} + θ_{2}}{2})$ $\Rightarrow \frac{2 m_{0}}{1 - m_{0}^{2}} = \frac{m_{1} + m_{2}}{1 - m_{1} m_{2}} . . . (i)$ $m_{1} m_{2} m_{0} = - a / d . . . . (i i)$ $(m_{1} + m_{2}) m_{0} + m_{1} m_{2} = b / d . . . (i i i)$ $m_{1} + m_{2} + m_{0} = - c / d . . . . (i v)$ $u s i n g (i i) a n d (i v) i n (i) a n d (i i i)$ $t o o b t a i n t w o e q u a t i o n s i n m_{0} :$ $\frac{2 m_{0}}{1 - m_{0}^{2}} = - \frac{(m_{0} + \frac{c}{d})}{(1 + \frac{a}{m_{0} d})}$ $\Rightarrow \frac{2}{1 - m_{0}^{2}} = - \frac{c + m_{0} d}{a + m_{0} d} . . . . . (I)$ $m_{0} = - \frac{(\frac{b}{d} + \frac{a}{m_{0} d})}{(m_{0} + \frac{c}{d})}$ $\Rightarrow m_{0}^{2} = - \frac{a + m_{0} b}{c + m_{0} d} . . . . . (I I)$ $(I) \times (I I) g i v e s$ $\frac{2 m_{0}^{2}}{1 - m_{0}^{2}} = \frac{a + m_{0} b}{a + m_{0} d}$ $a d d i n g 2 o n b o t h s i d e s$ $\frac{2}{1 - m_{0}^{2}} = \frac{3 a + m_{0} (b + 2 d)}{a + m_{0} d} . . . . . (A)$ $e q u a t i n g (I) w i t h (A) :$ $\frac{2}{1 - m_{0}^{2}} = \frac{3 a + m_{0} (b + 2 d)}{a + m_{0} d} = - \frac{c + m_{0} d}{a + m_{0} d}$ $h e n c e$ $3 a + m_{0} (b + 2 d) = - (c + m_{0} d)$ $m_{0} = - \frac{3 a + c}{b + 3 d}$ $s u b s t i t u t i n g i n (I I) :$ ${(\frac{3 a + c}{b + 3 d})}^{2} = - \frac{a - (\frac{3 a + c}{b + 3 d}) b}{c - (\frac{3 a + c}{b + 3 d}) d}$ $o r$ ${(\frac{3 a + c}{b + 3 d})}^{2} = - \frac{3 a d - 2 a b - b c}{b c + 2 c d - 3 a d}$ $\Rightarrow {(\frac{3 a + c}{b + 3 d})}^{2} = \frac{b c + 2 a b - 3 a d}{b c + 2 c d - 3 a d} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum