# Condition that the General Equation of Second Degree may Represent a Line Pair

The general equation of second degree in $x$ and $y$ is, $ax^2+2hxy+by^2+2gx+2fy+c=0\text{ }(a≠0)$ $\text{or, }ax^2+(2hy+2g)x+(by^2+2fy+c)=0$ which is quadratic in $x$. Solving for $x$, $x=\frac{-(2hy+2g)\pm\sqrt{(2hy+2g)^2-4a(by^2+2fy+c}}{2a}$ $ax=-(hy+g)\sqrt{(hy+g)^2-a(by^2+2fy+c)}$ $ax=-(hy+g)\pm\sqrt{h^2y^2+2ghy+g^2-aby^2-2afy-ac}$ $ax=-(hy+g)\pm\sqrt{(h^2-ab)y^2+(2gh-2af)y+(g^2-ac)}$

The general equation of second degree in $x$ and $y$ will represent a line pair if it can be resolved into two linear factors. For this, the quantity under the radical sign must be a perfect square. $\text{i.e. }B^2-4AC=0$ $(2gh-2af)^2-4(h^2-ab)(g^2-ac)=0$ $(gh-af)^2-(h^2-ab)(g^2-ac)=0$ $-a(abc+2fgh-af^2-bg^2-ch^2)=0$ Since $a≠0$, the required condition is $abc+2fgh-af^2-bg^2-ch^2=0$ If $a=0$ but $b≠0$, then the given equation can be expressed as a quadratic in $y$ and we can find the condition in the same way.

This condition may also be expressed in the determinant form as $\left|\begin{array}{c}a&h&g\\h&b&f\\g&f&c\end{array}\right|=0$ If $a=b=0$ and $h≠0$, then the given equation will reduce to $2hxy+2gx+2fy+c=0$ This can be written as $2x(hy+g)+\frac{2f}{h}(hy+g)-\frac{2fg}{h}+c=0$ $2(hy+g)\left(x+\frac{f}{h}\right)-\frac{2fg}{h}+c=0$ This represents two straight lines if $-\frac{2fg}{h}+c=0$ $\text{or, }2fg-ch=0$

### Prove that the equation $6x^2-xy-12y^2-8x+29y-14=0$ represents a pair of lines.

Given equation is, $6x^2-xy-12y^2-8x+29y-14=0$ $\begin{array}{c}\therefore a=6, &2h=-1, &b=-12 \\ a=6, &h=-\frac{1}{2}, &b=-12\end{array}$ $\begin{array}{c} \text{and, } 2g=-8, &2f=29, &c=-14\\ g=-4, &f=\frac{29}{2}, &c=-14\end{array}$ Now, the above equation will represent a line pair if $abc+2fgh-af^2-bg^2-ch^2=0$ Now, $abc+2fgh-af^2-bg^2-ch^2$ $=1008+58-\frac{2523}{8}+192+\frac{7}{2}$ $=0$ Hence, the given equation represents a line pair.

### Find the value of $k$ so that the equation $2x^2+7xy+3y^2-4x-7y+k=0$ may represent a line pair.

Given equation is, $2x^2+7xy+3y^2-4x-7y+k=0$ $\begin{array}{c}\therefore a=2, &2h=7, &b=3 \\ a=2, &h=\frac{7}{2}, &b=3\end{array}$ $\begin{array}{c} \text{and, } 2g=-4, &2f=-7, &c=k\\ g=-2, &f=-\frac{7}{2}, &c=k\end{array}$ The given equation will represent a line pair if $abc+2fgh-af^2-bg^2-ch^2=0$ $6k+49-\frac{49}{2}-12-\frac{49k}{4}=0$ $24k-196-98-48-49k=0$ $\therefore k=2$