# 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

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:

\(MP\) (angle bisector which lies on the side of 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

\(MQ\) (angle bisector which doesn't lies on the side of 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,

\[ RL_1 = \frac{|a_1h+b_1k+c_1|}{\sqrt{a^{2}_1+b^{2}_1}}\] \[ RL_2 = \frac{|a_2h+b_2k+c_2|}{\sqrt{a^{2}_2+b^{2}_2}}\]

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 and 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 for a pair of lines \[ax^2+2hxy+by^2=0 \], \[ \frac{a}{b} x^2 + \frac{2h}{b} xy + y^2 = 0 \] 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 x-axis respectively by \(Φ \) and \( Φ'\). For two angle bisector, one external and one internal, they are perpendicular to each other. So the angles between internal ,external angle bisectors and x-axis, can be expressed by \( \frac{Φ'+Φ}{2} \) and \( \frac{Φ'+Φ+π}{2} \) respectively. \[ tan (Φ'+Φ) = tan (Φ'+Φ+π) = \frac {m+m'}{1-mm'} \] \[ tan (Φ'+Φ) = \frac {-2h}{b-a} .......① \] Suppose there is a point \( (x,y) \) on one of the angle bisector, \[ tan (Φ'+Φ) = \frac {2( \frac{y}{x} ) }{1- ( \frac{y^2}{x^2} ) } \] \[ tan (Φ'+Φ) = \frac {2xy}{x^2-y^2} ....... ② \] Equating ① and ② , \[ \frac {2xy}{x^2-y^2} = \frac {-2h}{b-a} \] \[ \frac {xy}{h} = \frac {x^2-y^2}{a-b} \] \[ hx^2 - (a-b)xy - hy^2 = 0 \]

## Generalisation of Equation of two Angle Bisectors

But what if the two angle bisector 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, the coordinates of a point is now \( (X,Y) \) , which is \( (x-p, y-q) \). On substituting, we get \[ hX^2 - (a-b)XY - hY^2 = 0 \] \[ h(x-p) - (a-b)(x-p)(y-q) - h(y-q)^2 = 0 \] 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/