The question states the following:-

The bisector of two lines $L_1$ and $L_2$ are given by $3x^2-8xy-3y^2+10x+20y-25 = 0$. If the line $L_1$ passes through the origin, find the equation of $L_2$.

Now I know that the two angle bisectors are perpendicular to each other and I also know the formula to find the angle bisectors of two given lines if the equations of the lines themselves are known, but how do I get the equations of the lines back when only the equations of the bisectors are given?

I can also find the point of intersection of the two bisectors from the combined equation I guess, but what clue would that give me?

I really have no way to approach this problem.

  • $\begingroup$ Line $L_1$ passes through $(0,0)$ and through the intersection point of the bisectors, hence it can be found. $\endgroup$ – Intelligenti pauca Mar 25 '20 at 15:09
  • $\begingroup$ @Aretino Yes I did get the answer. Had to factor the expression, get the point of intersection and get line L$_1$ and then had to find L$_1$ by using that fact that L$_2$ and L$_1$ are equally inclined to a bisector $\endgroup$ – Techie5879 Mar 25 '20 at 15:11
  • $\begingroup$ You don’t need to factor the entire expression; just factoring the quadratic part is sufficient to get the directions of the bisectors. If you can’t readily factor that, you can get their direction vectors by solving $3x^2-8xy-3y^2=0$ for one of the variables after setting the other equal to $1$. However, if you’ve factored the equation completely, you can save yourself a bit of work by recognizing that $L_1$ is an angle bisector of the bisectors. $\endgroup$ – amd Mar 25 '20 at 20:34

You have to note that the intersection of pair of straight lines is same as that of pair of it's angle bisectors.

So, first you need to find the intersection of angle bisectors $(x_0,y_0)$ by $$\frac{\partial(L_1'L_2')}{\partial x}=0\ \text{and}\ \frac{\partial(L_1'L_2')}{\partial y}=0$$

Now, the equation of pair of angle bisectors is given by $$\frac{(x-x_0)^2-(y-y_0)^2}{a-b}=\frac{(x-x_0)(y-y_0)}{h}$$

This equation is same as the given equation $L_1'L_2'$. Compare the co-efficients to get the value of $a, b, h$ (ultimately the slopes of both the lines) as you have found $(x_0, y_0)$.

$(x_0, y_0)\equiv(1, 2) \\ L_1L_2\equiv 2x^2+3xy-2y^2-10x+5y=0$



In 2-D, if you take the unit parallel vectors of two lines ${\bf t}_1, \, {\bf t}_2$, then $$ {\bf u}_1 = {\bf t}_1 + {\bf t}_2 , \;{\bf u}_2 = {\bf t}_1 - {\bf t}_2$$ are vectors parallel to the two bisectors.


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