Show that the line y=mx bisects the angle between lines $a x^{2} - 2 h x y + b y^{2} = 0$ $ax^2-2hxy+by^2=0$ if $h (1 - m^{2}) + m (a - b) = 0$ $h(1-m^2)+m(a-b)=0$ ?

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Show that the line y=mx bisects the angle between lines $a x^{2} - 2 h x y + b y^{2} = 0$ $ax^2-2hxy+by^2=0$ if $h (1 - m^{2}) + m (a - b) = 0$ $h(1-m^2)+m(a-b)=0$ ?

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Answer 1 · 2018-08-06T15:52:58

Let the equations $y - m_{1} x = 0 and y - m_{2} x = 0$ $y-m_1x=0andy-m_2x=0$ are two straight lines represented by the given equation of pair of straight lines.Here $m_{1} = tan α and m_{2} = tan β and β > α$ $m_1=tanalphaandm_2=tanbetaand beta>alpha$

Hence

$(y - m_{1}) (y - m_{2} x) = y^{2} - \frac{2 h}{b} x y + \frac{a}{b} x^{2}$ $(y-m_1)(y-m_2x)=y^2-(2h)/bxy+a/bx^2$

So

$m_{1} + m_{2} = = \frac{2 h}{b} and m_{1} m_{2} = \frac{a}{b}$ $m_1+m_2==(2h)/bandm_1m_2=a/b$

If $θ$ $theta$ be the angle subtended by angle bisector ( $y = m x$ $y=mx$ ) of the pair of straight line with the positive direction of X-axis ,then $m = tan θ$ $m=tantheta$

Now it is obvious that

$θ - α = β - θ$ $theta-alpha=beta-theta$

So

$α + β = 2 θ$ $alpha+beta=2theta$

$\Rightarrow tan (α + β) = tan (2 θ)$ $=>tan(alpha+beta)=tan(2theta)$

$\Rightarrow \frac{tan α + tan β}{1 - tan α tan β} = \frac{2 tan θ}{1 - {tan}^{2} θ}$ $=>(tanalpha+tanbeta)/(1-tanalphatanbeta)=(2tantheta)/(1-tan^2theta)$

$\Rightarrow \frac{m_{1} + m_{2}}{1 - m_{1} m_{2}} = \frac{2 m}{1 - m^{2}}$ $=>(m_1+m_2)/(1-m_1m_2)=(2m)/(1-m^2)$

$\Rightarrow \frac{\frac{2 h}{b}}{1 - \frac{a}{b}} = \frac{2 m}{1 - m^{2}}$ $=>((2h)/b)/(1-a/b)=(2m)/(1-m^2)$

$\Rightarrow \frac{h}{b - a} = \frac{m}{1 - m^{2}}$ $=>h/(b-a)=m/(1-m^2)$

$\Rightarrow h (1 - m^{2}) = (b - a) m$ $=>h(1-m^2)=(b-a)m$

$\Rightarrow h (1 - m^{2}) + (a - b) m = 0$ $=>h(1-m^2)+(a-b)m=0$

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Show that the line y=mx bisects the angle between lines $a x^{2} - 2 h x y + b y^{2} = 0$ $ax^2-2hxy+by^2=0$ if $h (1 - m^{2}) + m (a - b) = 0$ $h(1-m^2)+m(a-b)=0$ ?

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Show that the line y=mx bisects the angle between lines $a x^{2} - 2 h x y + b y^{2} = 0$ $ax^2-2hxy+by^2=0$ if $h (1 - m^{2}) + m (a - b) = 0$ $h(1-m^2)+m(a-b)=0$ ?