Suppose you have a pair of lines passing through origin, $ax^2 + 2hxy +by^2 = 0$, how would you find the equation of pair of angle bisectors for this pair of lines. I can do this for $2$ separate lines, but I am not able to figure it out for a pair of lines. Can someone please help.
[Math] Equation of Angle Bisectors for a Pair of Straight Lines passing through the Origin
analytic geometry
Related Solutions
If the conic represents a pair of lines, then it can be written as $(px+qy+r)(p'x+q'y+r')$. Therefore, we get $ab = pp'qq'$ and $h = \frac{pq'+p'q}{2}$.
1) If both $pq'$ and $p'q$ are greater than equal to zero then apply AM-GM to $pq'$ and $p'q$ to get
$$\frac{pq'+ p'q}{2} \geq \sqrt{pq' \cdot p'q}$$
Squaring both sides, we get
$$h^2 = {\bigg(\frac{pq'+ p'q}{2}\bigg)}^2 \geq pp'qq' = ab$$
2) If exactly one of $pq'$ and $p'q$ is less than zero then $ab \leq 0$ and therefore is always less than equal to $h^2 \geq 0$.
3) If both $pq'$ and $p'q$ are less than zero then apply AM-GM to $-pq'$ and $-p'q$ to get
$$\frac{-pq'+ (-p'q)}{2} \geq \sqrt{-pq' \cdot -p'q} = \sqrt{pp'qq'}$$
Squaring both sides, we get
$$h^2 = {\bigg(\frac{-pq'+ (-p'q)}{2}\bigg)}^2 \geq pp'qq' = ab$$
The other two conditions come out in a similar manner.
In general, the equation $\ ax^2+2hxy+by^2+2gx+2fy+c=0\ $ defines a conic, of which two intersecting straight lines is one (degenerate) special case. One standard way to determine what form of conic the equation represents is to diagonalise the matrix, $$ A=\pmatrix{a&h\\h&b}\ $$ by finding its eigenvalues and egenvectors. Let $\ \lambda_1, \lambda_2\ $ be the eigenvalues, and $\ \boldsymbol{e}_1, \boldsymbol{e}_2\ $ the corresponding normalised eigenvectors, which can be chosen so that $\ \boldsymbol{e}_1=\pmatrix{\cos\theta\\-\sin\theta}\ $ and $\ \boldsymbol{e}_2=\pmatrix{\sin\theta\\\cos\theta}\ $ for some angle $\ \theta\ $. If $\ \Theta\ $ is the matrix with columns $\ \boldsymbol{e}_1\ $ and $\ \boldsymbol{e}_2\ $, then $\ \Theta\ $ is a rotation matrix, with $\ \Theta^{-1} = \Theta^\top\ $, and $$ \Theta^\top A\Theta = \pmatrix{\lambda_1 & 0\\0&\lambda_2}\ . $$ Now, the equation we are interested in can be written as \begin{eqnarray} 0&=& \boldsymbol{x}^\top A\boldsymbol{x} + 2\boldsymbol{g}^\top\boldsymbol{x} + c\\ &=& \left(\Theta^\top \boldsymbol{x}\right)^\top\Theta^\top A\Theta\left(\Theta^\top\boldsymbol{x}\right) + 2\left(\Theta^\top\boldsymbol{g}\right)^\top \Theta^\top\boldsymbol{x}+c\\ &=&\boldsymbol{x}'^\top\Lambda\boldsymbol{x}'+2\boldsymbol{g}'^\top\boldsymbol{x}'+c\ , \end{eqnarray} where $\ \boldsymbol{x}=\pmatrix{x\\y}\ $, $\ \boldsymbol{x}'=\Theta^\top\boldsymbol{x}\\$, $\ \Lambda = \pmatrix{\lambda_1 & 0\\0&\lambda_2}\ $, $\ \boldsymbol{g}=\pmatrix{g\\h}\ $, and $\ \boldsymbol{g}'=\Theta^\top\boldsymbol{g}\ $. The entries of $\ \boldsymbol{x}'\ $, $\ x_1'=x\cos\theta+y\sin\theta\ $, and $\ x_2'=-x\sin\theta+y\cos\theta\ $, are the coordinates of a point $P$ with respect to a set of axes that have been rotated clockwise through an angle $\ \theta\ $relative to the original axes, where $\ x\ $ and $ y\ $ are the coordinates of $P$ with respect to those original axes. It follows from above that the equation the the conic with respect to the new axes is \begin{eqnarray} 0 &=& \lambda_1 x_1'^2 + \lambda_2x_2'^2 +2g_1'x_1'+ 2g_2'x_2' + c\\ &=& \lambda_1\left(x_1' +\frac{g_1'}{\lambda_1}\right)^2 + \lambda_2\left(x_2' +\frac{g_2'}{\lambda_2}\right)^2 +c - \frac{g_1'^2}{\lambda_1}-\frac{g_2'^2}{\lambda_2}\ . \end{eqnarray} This is the equation of two intersecting straight lines if and only if $\ \lambda_1\ne0\ $, $\ \lambda_2\ne0\ $, $\ \lambda_1\ $ and $\ \lambda_2\ $ are of opposite sign, and $\ c - \frac{g_1'^2}{\lambda_1}-\frac{g_2'^2}{\lambda_2}=0\ $. If this is the case, suppose, without loss of generality, that $\ \lambda_1>0\ $ and $\ \lambda_2<0\ $. Then the above equation becomes \begin{eqnarray} 0 &=& \left(\sqrt{\lambda_1}x_1' + \frac{g_1'}{\sqrt{\lambda_1}}\right)^2- \left(\sqrt{-\lambda_2}x_2' - \frac{g_2'}{\sqrt{-\lambda_2}}\right)^2\\ &=& \left(\sqrt{\lambda_1}x_1' + \sqrt{-\lambda_2}x_2'+ \frac{g_1'}{\sqrt{\lambda_1}}-\frac{g_2'}{\sqrt{-\lambda_2}}\right)\\ &&\ \ \ \cdot \left(\sqrt{\lambda_1}x_1' - \sqrt{-\lambda_2}x_2'+ \frac{g_1'}{\sqrt{\lambda_1}}+\frac{g_2'}{\sqrt{-\lambda_2}}\right)\ , \end{eqnarray} and the equations of the two straight lines in the new coordinates are \begin{eqnarray} \sqrt{\lambda_1}x_1' + \sqrt{-\lambda_2}x_2'+ \frac{g_1'}{\sqrt{\lambda_1}}-\frac{g_2'}{\sqrt{-\lambda_2}}&=&0\ \ \mbox{, and}\\ \sqrt{\lambda_1}x_1' - \sqrt{-\lambda_2}x_2'+ \frac{g_1'}{\sqrt{\lambda_1}}+\frac{g_2'}{\sqrt{-\lambda_2}} &=&0\ . \end{eqnarray} Finally, substituting $\ x_1'=x\cos\theta+y\sin\theta\ $, and $\ x_2'=-x\sin\theta+y\cos\theta\ $ in these equations, we get the equations of the lines in the original coordinates: \begin{eqnarray} x\left(\sqrt{\lambda_1}\cos\theta - \sqrt{-\lambda_2}\sin\theta\right)&+ &y\left(\sqrt{\lambda_1}\sin\theta +\sqrt{-\lambda_2}\cos\theta\right)\\ &+& \frac{g_1'}{\sqrt{\lambda_1}}-\frac{g_2'}{\sqrt{-\lambda_2}}&=&0 \end{eqnarray} and \begin{eqnarray} x\left(\sqrt{\lambda_1}\cos\theta + \sqrt{-\lambda_2}\sin\theta\right)&+ &y\left(\sqrt{\lambda_1}\sin\theta -\sqrt{-\lambda_2}\cos\theta\right)\\ &+& \frac{g_1'}{\sqrt{\lambda_1}}+\frac{g_2'}{\sqrt{-\lambda_2}}&=&0\ . \end{eqnarray}
Best Answer
Let's consider an easier case, when the given lines have the form $y=mx$ and $y=nx$ (with $m\ne n$, of course).
The points on the angle bisectors satisfy the conditions that their distance from the two lines is the same. So you have, for such a point $(x,y)$, $$ \frac{|y-mx|}{\sqrt{1+m^2}}=\frac{|y-nx|}{\sqrt{1+n^2}} $$ By squaring, we get $$ (y-mx)^2(1+n^2)=(y-nx)^2(1+m^2) $$ and expanding $$ (y-mx)^2-(y-nx)^2=m^2(y-nx)^2-n^2(y-mx)^2. $$ The left hand side becomes $$ (y-mx+y-nx)(y-mx-y+nx)=-(m-n)(2y-(m+n)x)x $$ and the right hand side is $$ (my-mnx+ny-mnx)(my-mnx-ny+mnx)=(m-n)((m+n)y-2mnx)y $$ Since $m\ne n$, we can factor out $m-n$ on both sides, getting $$ -2xy+(m+n)x^2=(m+n)y^2-2mnxy $$ that can be written $$ (m+n)x^2-2(1-mn)xy-(m+n)y^2=0 $$ If you consider the equation $by^2+2hxy+ax^2=0$ in the unknown $y$, you can see that $m$ and $n$ are the roots of it (when $b\ne 0$). Thus $m+n=-2h/b$ and $mn=a/b$, so you can rewrite the above equation in the form $$ \frac{-2h}{b}x^2-2\left(1-\frac{a}{b}\right)xy-\frac{-2h}{b}y^2=0 $$ which is $$ hx^2-(a-b)xy-hy^2=0 $$ Note that it can't be both $h=0$ and $a=b\ne0$, because in this case the original equations would be $x^2+y^2=0$ (the isotropic lines, if you consider the complex plane).