# Angle between the line pair represented by ax²+2hxy+by²=0

Let the line pair be represented by the equation $ax^2+2hxy+by^2=0$ Let $y=m_1x$ and $y=m_2x$ be the equations of two lines represented by the above line pair where $m_1+m_2=-\frac{2h}{b}\text{ and }m_1m_2=\frac{a}{b}$ Let $\theta$ be the angle between the line pair. Then, $\tan\theta=\pm\frac{m_1-m_2}{1+m_1m_2}$ $=\pm\frac{\sqrt{(m_1+m_2)^2-4m_1m_2}}{1+m_1m_2}$ $=\pm\frac{\sqrt{\frac{4h^2}{b^2}-\frac{4a}{b}}}{1+\frac{a}{b}}=\pm\frac{2\sqrt{h^2-ab}}{a+b}$ $\therefore\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$

The homogeneous equation of degree two i.e. $ax^2+2hxy+by^2=0$ represents

1. two real and distinct lines if $h^2-ab>0$
2. two real and coincident lines if $h^2-ab=0$
3. two imaginary lines if $h^2-ab<0$

## Condition of Parallelism

Two lines will be parallel to each other if $\theta=0°$. $\therefore\tan 0°=\pm\frac{2\sqrt{h^2-ab}}{a+b}$ $\sqrt{h^2-ab}=0$ $\therefore h^2=ab$

## Condition of Perpendicularity

Two lines will be perpendicular to each other if $\theta=90°$. $\therefore\tan 90°=\pm\frac{2\sqrt{h^2-ab}}{a+b}$ $\cot 90°=\pm\frac{a+b}{2\sqrt{h^2-ab}}$ $\therefore a+b=0$ $\therefore\text{coeff. of }x^2+\text{coeff. of }y^2=0$

### Find the angle between the line pair $x^2+9xy+14y^2=0$.

Here, $x^2+9xy+14y^2=0$ $\begin{array}{c} \therefore a=1, & 2h=9 &\text{and}& b=14 \\ a=1, & h=\frac{9}{2} &\text{and}& b=14\end{array}$ Let $\theta$ be the angle between the line pair. Then, $\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$ $=\pm\frac{2\sqrt{\frac{81}{4}-14}}{1+14}=\pm\frac{2×5}{2×15}=\pm\frac{1}{3}$ $\therefore\theta=\tan^{-1}\frac{1}{3}$

### Find the angle between the line pair $x^2-5xy+4y^2+x+2y-2=0$.

Given line pair is, $x^2-5xy+4y^2+x+2y-2=0$ Taking homogeneous part only, $x^2-5xy+4y^2=0$ $\begin{array}{c} \therefore a=1, & 2h=-5 &\text{and}& b=4 \\ a=1, & h=\frac{-5}{2} &\text{and}& b=4\end{array}$ Let $\theta$ be the angle between the line pair. Then, $\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$ $=\pm\frac{2\sqrt{\frac{25}{4}-4}}{1+4}=\pm\frac{2×\frac{3}{2}}{5}=\pm\frac{3}{5}$ $\therefore\theta=\tan^{-1}\frac{3}{5}$

### Prove that the two lines are parallel in the line pair $x^2+6xy+9y^2+12y-5=0$.

Given line pair is, $x^2+6xy+9y^2+4x+12y-5=0$ Taking homogeneous part only, $x^2+6xy+9y^2=0$ $\begin{array}{c} \therefore a=1, & 2h=6 &\text{and}& b=9 \\ a=1, & h=3 &\text{and}& b=9\end{array}$ The two lines will be parallel if $h^2=ab$ $3^2=1×9$ $9=9$ Hence, the two lines are parallel.