How do I find the equation of lines if the combined equation of bisectors is given?

Question

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 $L_2$.

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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.

Answer 1

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 $(x_0, y_0)$.

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

Answer 2

Hint:

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.