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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}} \dots (i)

m_{1} m_{2} m_{0} = - a / d \dots . (i i)

(m_{1} + m_{2}) m_{0} + m_{1} m_{2} = b / d \dots (i i i)

m_{1} + m_{2} + m_{0} = - c / d \dots . (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} \dots . . (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} \dots . . (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} \dots . . (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} .

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