Excenters of a Triangle

This math recipe will help you find the excenters of a triangle, coordinates of whose vertices are known.

Easy

Approx. 5 min

Coordinates of the three vertices: \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\)

To find the excenters \(I_1\), \(I_2\), and \(I_3\) of \(\Delta ABC\), we’ll use the formula below:

\(I_1 \equiv \left ( \frac{-ax_1+bx_2+cx_3}{-a+b+c},\frac{-ay_1+by_2+cy_3}{-a+b+c} \right ) \)

\(I_2 \equiv \left ( \frac{ax_1-bx_2+cx_3}{a-b+c},\frac{ay_1-by_2+cy_3}{a-b+c} \right ) \)

\(I_3 \equiv \left ( \frac{ax_1+bx_2-cx_3}{a+b-c},\frac{ay_1+by_2-cy_3}{a+b-c} \right ) \)

Here, \(a\), \(b\), and \(c\) are the lengths of the sides \(BC\), \(CA\), and \(AB\). We’ll have to find these using the Distance Formula.

Example 1 Find the three excenters of the triangle, whose vertices are \(A(0, 0)\), \(B(3, 0)\), and \(C(0, 4)\).

Solution The figure shows \(\Delta ABC\) on the plane.

To find the excenters, we’ll first find the three sides \(a\), \(b\), and \(c\) using the Distance Formula:

\(a = BC = \sqrt{(3-0)^2+(0-4)^2} = 5\)

\(b = CA = \sqrt{(0-0)^2+(0-4)^2} = 4\)

\(c = AB = \sqrt{(3-0)^2+(0-0)^2} = 3\)

Now, we’ll use the formula for the excenters:

\(I_1 \equiv \left ( \frac{-5(0)+4(3)+3(0)}{-5+4+3},\frac{-5(0)+4(0)+3(4)}{-5+4+3} \right ) \)

\( \equiv \left ( \frac{12}{2},\frac{2}{12} \right ) \)

\( \equiv (6,6) \)

Let’s find the next one.

\(I_2 \equiv \left ( \frac{5(0)-4(3)+3(0)}{5-4+3},\frac{5(0)-4(0)+3(4)}{5-4+3} \right ) \)

\( \equiv \left ( \frac{-12}{4},\frac{12}{4} \right ) \)

\( \equiv (-3,3) \)

One more to go.

\(I_3 \equiv \left ( \frac{5(0)+4(3)-3(0)}{5+4-3},\frac{5(0)+4(0)-3(4)}{5+4-3} \right ) \)

\( \equiv \left ( \frac{12}{6},\frac{-12}{6} \right ) \)

\( \equiv (2,-2) \)

That’s it for this recipe. Hope you found it helpful.

For more recipes, please visit www.doubleroot.in/recipes.

You can follow me on Instagram, Twitter, or Facebook to get all updates.