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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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.

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,

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

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.

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,

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

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,

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

Equating \((1)\) and \((2)\) gives

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

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

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

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.

• Angle bisector theorem
• Distance between point and line
• Incenter