# Line Pair ax²+2hxy+by²=0 parallel to the General Equation

### If the Equations $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a Pair of Lines, then $ax^2+2hxy+by^2=0$ represent a Pair of Lines through the Origin Parallel to the above Pair.

Let $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a line pair. Then the left hand side can be resolved into two linear factors and the equation may be written as $(l_1x+m_1y+n_1)(l_2x+m_2y+n_2)=0$ where $l_1l_2=a$, $m_1m_2=b$, $n_1n_2=c$, $l_1m_2+m_1l_2=2h$, $l_1n_2+n_1l_2=2g$, $m_1n_2+m_2n_1=2f$.

The separate equations of the lines are $l_1x+m_1y+n_1=0\text{ and }l_2x+m_2y+n_2=0$ Hence the equations of the lines through the origin parallel to the above lines are $l_1x+m_1y=0$ and $l_2x+m_2y=0$. Then, the combined equation is $(l_1x+m_1y)(l_2x+m_2y)=0$ $l_1l_2x^2+(l_1m_2+l_2m_1)xy+m_1m_2y^2=0$ $\therefore ax^2+2hxy+by^2=0$

#### Angle between two lines represented by $ax^2+2hxy+by^2+2gx+2fy+c=0\text{ __(1)}$ is same as the angle between the two lines represented by $ax^2+2hxy+by^2=0\text{ __(2)}$

Since the two lines represented by the equation $\text{(2)}$ are parallel to the two lines represented by the equation $\text{(1)}$. So, the angle between the two lines given by $\text{(1)}$ is same as the angle between the lines given by $\text{(2)}$.

But the angle between the lines represented by $\text{(2)}$ is given by, $\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$ So, the angle between the two lines represented by equation $\text{(1)}$ is also given by the same formula $\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$ Hence, the two lines represented by $ax^2+2hxy+by^2+2gx+2fy+c=0$ will be perpendicular to each other if $a+b=0$ and they will be parallel to each other is $h^2=ab$. [From: Angle between the line pair represented by $ax^2+2hxy+by^2=0$]

The point of intersection of the two lines represented by the general equation of second degree can be obtained by solving the separate equations $l_1x+m_1y+n_1=0\text{ and }l_2x+m_2y+n_2=0$ for $x$ and $y$, and writing down the result in terms of the coefficients of the given equation by using the relations $l_1l_2=a$, $m_1m_2=b$, $n_1n_2=c$, $l_1m_2+m_1l_2=2h$, $l_1n_2+n_1l_2=2g$, $m_1n_2+m_2n_1=2f$. Then, the coordinates of the point of intersection will be $\left(\sqrt{\frac{f^2-bc}{h^2-ab}},\sqrt{\frac{g^2-ca}{h^2-ab}}\right)$