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 \[abc+2fgh-af^2-bg^2-ch^2=0\] \[6k+49-\frac{49}{2}-12-\frac{49k}{4}=0\] \[24k-196-98-48-49k=0\] \[\therefore k=2\]