# Thread: Angle bisector of paired lines

1. ## Angle bisector of paired lines

By considering the set of points P(x, y) that are equidistant from the two lines represented by the equation $ax^2+2hxy+by^2=0$ show that the equation of the pair of angle bisectors of these lines is
$hx^2+(b-a)xy-hy^2=0$

I know that the pair of lines that are bisectors are perpendicular to each other and all the lines intersect at the origin. I don't know what else, or how to begin.
Thanks!

2. The pair of line can be written as

$y - m_1x = 0$ and $y - m_2x = 0$

where $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$

Equations of angle bisectors are given by

$\frac{y - m_1x}{\sqrt{1+m_1^2}} - \frac{y - m_2x}{\sqrt{1+m_2^2}}= 0$

$\frac{y - m_1x}{\sqrt{1+m_1^2}} + \frac{y - m_2x}{\sqrt{1+m_2^2}}= 0$

The combined equation of the angle bisectors is given by

$\frac{(y - m_1x)^2}{1+m_1^2} - \frac{(y - m_2x)^2}{1+m_2^2}= 0$

Simplify and substitute the values of m1+m2 and m1m2.

3. umm how did you get the two first expressions? i don't understand

4. Originally Posted by arze
umm how did you get the two first expressions? i don't understand
Equation of a straight line passing through the origin is given by

y = mx, or y - mx = 0. If m1 and m2 are the slopes of a pair of straight lines, the combined equation is given by

$(y - m_1x)(y-m_2x) = 0$

$y^2 - (m_1+m_2)x + m_1m_2x^2 = 0$

Compare this equation with equation given in the problem. You can see that

$m_1 + m_2 = -\frac{2h}{b}$

and $m_1m_2 = \frac{a}{b}$

5. oops, sorry i meant these:
$\frac{y - m_1x}{\sqrt{1+m_1^2}} - \frac{y - m_2x}{\sqrt{1+m_2^2}}= 0$
$\frac{y - m_1x}{\sqrt{1+m_1^2}} + \frac{y - m_2x}{\sqrt{1+m_2^2}}= 0$
thanks!

6. Originally Posted by arze
oops, sorry i meant these:

thanks!
The angle bisector is the locus of points which are equidistant from the two straight lines.

The two equations are two angle bisectors obtained by finding the perpendicular distance from a point (x, y)