# Coordinate Geometry - Angle Bisector

In coordinate geometry, the equation of the angle bisector of two lines can be expressed in terms of those lines.

**Angle bisectors** are useful in constructing the incenter of a triangle, among other applications in geometry.

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## Equation of the Angle Bisector

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Let line $AB$ be defined by the equation $a_1x+b_1y+c_1=0$, and $CD$ be defined by the equation $a_2x+b_2y+c_2=0$. So long as these lines are not parallel lines (in which case the "angle bisector" does not exist), these two lines intersect at some point $M$.

Note that this defines two separate angles (but not four, as two pairs of vertical angles are equal): $\angle AMC$ and $\angle BMC$. Let $MP$ be the angle bisector of $\angle AMC$, and $MQ$ be the angle bisector of $\angle BMC$. Then,

$MP$ (the angle bisector which lies on the side of the origin) can be defined by the equation

$\frac{a_1x+b_1y+c_1}{\sqrt{a^{2}_1+b^{2}_1}} = \frac{a_2x+b_2y+c_2}{\sqrt{a^{2}_2+b^{2}_2}},$

and

$MQ$ (the angle bisector which doesn't lie on the side of the origin) can be defined as

$\frac{a_1x+b_1y+c_1}{\sqrt{a^{2}_1+b^{2}_1}} =-\frac{a_2x+b_2y+c_2}{\sqrt{a^{2}_2+b^{2}_2}}.$

Notice that these equations are practically identical; one is simply the "negative" of the other. This is not surprising, since the two angle bisectors are necessarily perpendicular lines.

## Proof

Let $MP$ be the angle bisector of $\angle AMC$, and let $R=(h,k)$ be a point on this bisector. Let $L_1$ and $L_2$ be the feet of the two perpendiculars from $R$ to $AB$ and $CD$, respectively. Then triangles $MRL_1$ and $MRL_2$ are congruent and equal in all respects. Hence, $RL_1=RL_2$.

By the distance between point and line formula,

$\begin{aligned} RL_1 &= \frac{|a_1h+b_1k+c_1|}{\sqrt{a^{2}_1+b^{2}_1}}\\\\ RL_2 &= \frac{|a_2h+b_2k+c_2|}{\sqrt{a^{2}_2+b^{2}_2}}. \end{aligned}$

WLOG, suppose that $MP$ lies on the same side of $AB,CD$ as the origin does $($else, swap $P$ and $Q)$. Note that this implies $a_1h+b_1k+c_1$ and $a_2h+b_2k+c_2k$ will have same sign as $c_1$ and $c_2$ will have, respectively. Assume that $c_1$ and $c_2$ have the same sign $($else, again swap $P$ and $Q)$, so that $a_1h+b_1k+c_1$ and $a_2h+b_2k+c_2$ both have the same sign. Thus

$\frac{a_1h+b_1k+c_1}{\sqrt{a^{2}_1+b^{2}_1}}=\frac{a_2h+b_2k+c_2}{\sqrt{a^{2}_2+b^{2}_2}}$

for all $R \in MP$. It is much easier to show that all $R$ satisfying the above lie on $MP$ by reversing the steps, so the equation of $MP$ is precisely the above, as desired.

In the alternative cases, the sign of $a_1h+b_1k+c_1$ and $a_2h+b_2k+c_2$ would be different, so

$\frac{a_1h+b_1k+c_1}{\sqrt{a^{2}_1+b^{2}_1}}=-\frac{a_2h+b_2k+c_2}{\sqrt{a^{2}_2+b^{2}_2}},$

which gives the equation of $MQ$, as desired.

## Equation of two Angle Bisectors

Suppose we have a pair of lines

$ax^2+2hxy+by^2=0 \quad \text{ or } \quad \frac{a}{b} x^2 + \frac{2h}{b} xy + y^2 = 0 \ \ (b\ne 0),$

which can be written in form of $(mx-y)(m'x-y) = 0$, where $mm' = \frac{a}{b}$ and $m+m' = \frac{-2h}{b}$. We denote the angles between $mx=y$ and $m'x=y$, and the $x$-axis, respectively, by $\Phi$ and $\Phi '$. The two angle bisectors, one external and one internal, are perpendicular to each other. So the angles between the internal and external angle bisectors and the $x$-axis can be expressed by $\frac{\Phi '+\Phi}{2}$ and $\frac{\Phi '+\Phi+\pi}{2}$, respectively.

Now, recall the following trigonometric identity:

$\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}.$

Using this relation,

$\begin{aligned} \tan (\Phi '+\Phi) &= \frac{\tan(\Phi) + \tan(\Phi')}{1 - \tan(\Phi)\tan(\Phi')} \\ &= \frac {m+m'}{1-mm'} \\ &= \frac {-2h}{b-a}. \qquad (1) \end{aligned}$

Suppose there is a point $(x,y)$ on one of the angle bisector in red below, and call the angle of the bisector with respect to the $x$-axis $\Phi_\text{bisector} = \frac{\Phi + \Phi'}{2}.$

Again, using the above trigonometric relation,

$\begin{aligned} \tan (\Phi '+\Phi) = \tan (\Phi_\text{bisector}+\Phi_\text{bisector}) &= \frac{\tan(\Phi_\text{bisector}) + \tan(\Phi_\text{bisector})}{1 - \tan({\Phi_\text{bisector}})\tan(\Phi_\text{bisector})} \\ &= \frac {2\left( \frac{y}{x} \right) }{1- \left( \frac{y^2}{x^2} \right) } \\ &= \frac {2xy}{x^2-y^2}. \qquad (2) \end{aligned}$

Equating $(1)$ and $(2)$ gives

$\begin{aligned} \frac {2xy}{x^2-y^2} &= \frac {-2h}{b-a} \\\\ \frac {xy}{h} &= \frac {x^2-y^2}{a-b} \\\\ hx^2 - (a-b)xy - hy^2 &= 0. \end{aligned}$

## Generalisation of Equation of Two Angle Bisectors

But what if the two angle bisectors don't intercept at the origin? This is a simple matter by using only the secondary equation of lines. First, relocate the origin at $(p,q)$, with respect to new origin, then the coordinates of a point are now $(X,Y)$, which is $(x-p, y-q)$. On substituting, we get

$\begin{aligned} hX^2 - (a-b)XY - hY^2 &= 0 \\ h(x-p)^2 - (a-b)(x-p)(y-q) - h(y-q)^2 &= 0. \end{aligned}$

Expanding it is left as an exercise for you. Remember that its expansion is in form of $ax^2 +2hxy + by^2 + 2gx + 2fy + c$. Check whether

$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$

is zero to confirm that the equation given is a pair of lines.

## See Also

**Cite as:**Coordinate Geometry - Angle Bisector.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/coordinate-geometry-angle-bisector/