Summary
This math recipe will help you find the incenter of a triangle, coordinates of whose vertices are known.
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.
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.
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.
For more recipes, please visit www.doubleroot.in/recipes.
You can follow me on Instagram, Twitter, or Facebook to get all updates.