Incenter of a Triangle

Summary

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

Incenter of a Triangle

Skill Level

Easy

Time

Approx. 5 min

Ingredients

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

Method

To find the incenter \(I\) of \(\Delta ABC\), we’ll simply plug their coordinates into the formula below:

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

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.

Examples

Example 1 Find the incenter 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.

Incenter of a Triangle

To find the incenter \(I\), 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 incenter:

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

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

\( \equiv (1,1) \)

Example 2 Find the incenter of the triangle, whose vertices are \(A(0, 0)\), \(B(0, -14)\), and \(C(12, -5)\).

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

Incenter of a Triangle

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

\(a = BC = \sqrt{(0-12)^2+(-14-(-5))^2} = 15\)

\(b = CA = \sqrt{(12-0)^2+(-5-0)^2} = 13\)

\(c = AB = \sqrt{(0-0)^2+(0-(-14))^2} = 14\)

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

\(I \equiv \left ( \frac{15(0)+13(0)+14(12)}{15+13+14},\frac{15(0)+13(-14)+14(-5)}{15+13+14} \right ) \)

\( \equiv \left ( \frac{168}{42},\frac{-252}{42} \right ) \)

\( \equiv (4,-6) \)

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

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