# Bisectors of the Angles between the Line Pair ax²+2hxy+by²=0

Let the two straight lines represented by the equation $ax^2+2hxy+by^2=0$ be $y=m_1x$ and $y=m_2x$ so that $m_1+m_2=-\frac{2h}{b}\text{ and }m_1m_2=\frac{a}{b}$ Let $OA$ and $OB$ be the two lines as in the figure and let $OC$ and $OD$ be the bisectors of the angles between them.

If $OA$ and $OB$ make angles $\theta_1$ and $\theta_2$ with $OX$, then $\tan\theta_1=m_1$ and $\tan\theta_2=m_2$. Let $OC$ make an angle $\phi$ with $OX$. From the figure, we have $\angle XOC=\angle XOA+\angle AOC$ $\phi=\theta_1+\frac{1}{2}(\theta_2-\theta_1)=\frac{1}{2}(\theta_1+\theta_2)$ $2\phi=\theta_1+\theta_2$ $\tan 2\phi=\tan(\theta_1+\theta_2)$ $\text{i.e. }\tan(2\angle XOC)=\tan 2\phi=\tan(\theta_1+\theta_2)$

Also, $\angle XOD=\angle XOC+90°=\phi+90°$ $\text{and }\tan(2\angle XOD)=\tan(2\phi+180°)=\tan 2\phi$ Hence if $(x,y)$ be any point in $OC$, or $OD$, we have $\tan\angle XOC$ or $\tan\angle XOD=\frac{y}{x}$, and $\tan 2\angle XOD=\tan 2\angle XOC=\tan 2\phi$.

$\text{But }\tan 2\phi=\tan(\theta_1+\theta_2)$ $\therefore\frac{2\tan\phi}{1-\tan^2\phi}=\frac{\tan\theta_1+\tan \theta_2}{1-\tan\theta_1\tan\theta_2}$ $\frac{2\frac{y}{x}}{1-\left(\frac{y}{x}\right)^2}=\frac{m_1+m_2}{1-m_1m_2}$ $\frac{2\frac{y}{x}}{1-\frac{y^2}{x^2}}=\frac{-\frac{2h}{b}}{1-\frac{a}{b}}$ $\therefore h(x^2-y^2)=(a-b)xy$ $\text{or, }\frac{xy}{x^2-y^2}=\frac{h}{a-b}$ Since this is true for any point $(x,y)$ in $OC$ or $CD$, it represents the two bisectors.

### Alternative Method

Here, $ax^2+2hxy+by^2=0$ Let $\begin{array}{c}y=m_1x &\text{and}& y=m_2x\\ \text{or, } m_1x-y=0 &\text{and}& m_2x-y=0\end{array}$ be the two lines represented by $ax^2+2hxy+by^2=0$. Then, $m_1+m_2=-\frac{2h}{b}\text{ and }m_1m_2=\frac{a}{b}$ Now, the equations of the bisectors of the angles between above lines are $\frac{m_1x-y}{\sqrt{m_1^2+1}}=\pm\frac{m_2x-y}{\sqrt{m_2^2+1}}$

$(m_1x-y)^2(1+m_2^2)=(m_2x-y)^2(1+m_1^2)$ $\text{or, }(m_1^2x^2-2m_1xy+y^2)(1+m_2^2)$$=(m_2^2x^2-2m_2xy+y^2)(1+m_1^2)$ $\text{or, }m_1^2(1+m_2^2)x^2-2m_1(1+m_2^2)xy+(1+m_2^2)y^2$ $=m_2^2(1+m_1^2)x^2-2m_2(1+m_1^2)xy+(1+m_1^2)y^2$ $\text{or, }[m_1^2(1+m_2^2)-m_2^2(1+m_1^2)]x^2+[(1+m_2^2)$$-(1+m_1^2)]y^2=2[m_1(1+m_2^2)-m_2(1+m_1^2)]xy$ $\text{or, }(m_1^2+m_1^2m_2^2-m_2^2-m_2^2m_1^2)x^2-(m_1^2-m_2^2)y^2$ $=2[m_1+m_1m_2^2-m_2-m_1^2m_2]xy$ $\text{or, }(m_1^2-m_2^2)x^2-(m_1^2-m_2^2)y^2$ $=2[m_1-m_2-m_1m_2(m_1-m_2)]xy$

$\text{or, }(m_1^2-m_2^2)(x^2-y^2)=2(m_1-m_2)(1-m_1m_2)xy$ $\text{or, }(m_1+m_2)(x^2-y^2)=2(1-m_1m_2)xy$ $\text{or, }\frac{m_1+m_2}{1-m_1m_2}=\frac{2xy}{x^2-y^2}$ $\text{or, }\frac{-\frac{2h}{b}}{1-\frac{a}{b}}=\frac{2xy}{x^2-y^2}$ $\text{or, }\frac{-2h}{b-a}=\frac{2xy}{x^2-y^2}$ $\text{or, }\frac{2h}{a-b}=\frac{2xy}{x^2-y^2}$ $\therefore\frac{xy}{x^2-y^2}=\frac{h}{a-b}$ which is the required equation.

### Find the equations of the bisectors of the angles between the line pair $3x^2-15xy+2y^2=0$.

Given line pair is, $3x^2-15xy+2y^2=0$ $\begin{array}{c}\therefore a=3, & 2h=-15 & \text{and} & b=2 \\ a=3, & h=-\frac{15}{2} &\text{and}& b=2\end{array}$ The equations of the bisectors are, $\frac{xy}{x^2-y^2}=\frac{h}{a-b}$ $\frac{xy}{x^2-y^2}=\frac{-\frac{15}{2}}{3+2}$ $\frac{xy}{x^2+y^2}=-\frac{3}{2}$ $2xy=-3x^2+3y^2$ $\therefore 3x^2+2xy-3y^2=0$

### If the line pairs $ax^2+2hxy+by^2=0$ and $a’x^2+2h’xy+b’y^2=0$ have the same bisectors, prove that $h(a’-b’)=h'(a-b)$.

Here, $ax^2+2hxy+by^2=0\text{ __(1)}$ The equations of the bisectors of the angles between the line pair $\text{(1)}$ are $\frac{xy}{x^2-y^2}=\frac{h}{a-b}$ Also, $a’x^2+2h’xy+b’y^2=0\text{ __(2)}$ The equations of the bisectors of the angles between the line pair $\text{(2)}$ are $\frac{xy}{x^2-y^2}=\frac{h’}{a’-b’}$ Since the bisectors are same, $\frac{h}{a-b}=\frac{h’}{a’-b’}$ $\therefore h(a’-b’)=h'(a-b)$

### If the pair of lines $x^2-2pxy-y^2=0$ and $x^2-2qxy-y^2=0$ be such that each pair bisects the angles between the other pair, prove that $pq=-1$.

The pair of bisectors of the angles between the line pair $x^2-2pxy-y^2=0$ is given by, $\frac{xy}{x^2-y^2}=\frac{-p}{1-(-1)}$ $2xy=-px^2+py^2$ $\therefore px^2+2xy-py^2=0$ According to the question, the pair of bisectors is given by $x^2-2qxy-y^2=0$. Hence, $px^2+2xy-py^2=0$ and $x^2-2qxy-y^2=0$ must represent the same pair of lines. $\therefore\frac{p}{1}=\frac{2}{-2q}=\frac{-p}{-1}$ $\therefore pq=-1$