Question

If $I_1,I_2,I_3$ are excentres of the triangle with vertices $\left(0,0\right),\left(5,12\right),\left(16,12\right)$ then find the orthocenter of $△I_1{I}_2{I}_3$

Answer 1

Step 1: Calculate the lengths of the sides of the triangle.

It is given that $△I_1{I}_2{I}_3$ are excenters of the triangle whose vertices are $\left(0,0\right),\left(5,12\right),\left(16,12\right)$.

First, draw a $△ABC$ with vertices $\left(0,0\right),\left(5,12\right),\left(16,12\right)$.

Now, according to the distance formula, the distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ and be calculated by,

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

So, the lengths of the sides $AB,BC$ and $CA$ are,

$AB=\sqrt{{\left(5-0\right)}^2+{\left(12-0\right)}^2}$

$$\begin{gathered} \Rightarrow AB=\sqrt{{\left(5\right)}^2+{\left(12\right)}^2} \\ \Rightarrow AB=\sqrt{25+144} \\ \Rightarrow AB=\sqrt{169} \\ \Rightarrow AB=13 \end{gathered}$$

And, $BC=\sqrt{{\left(16-5\right)}^2+{\left(12-12\right)}^2}$

$$\begin{gathered} \Rightarrow BC=\sqrt{{\left(11\right)}^2+{\left(0\right)}^2} \\ \Rightarrow BC=\sqrt{121+0} \\ \Rightarrow BC=\sqrt{121} \\ \Rightarrow BC=11 \end{gathered}$$

Also, $CA=\sqrt{{\left(0-16\right)}^2+{\left(0-12\right)}^2}$

$$\begin{gathered} \Rightarrow CA=\sqrt{{\left(-16\right)}^2+{\left(-12\right)}^2} \\ \Rightarrow CA=\sqrt{256+144} \\ \Rightarrow CA=\sqrt{400} \\ \Rightarrow CA=20 \end{gathered}$$

Step 2: Calculate the coordinates of the incenter

As we know, the coordinates of the incenter of a triangle with sides $a,b,c$ and vertices $\left(x_1,y_1\right),\left(x_2,y_2\right),\left(x_3,y_3\right)$ can be calculated by the formula,

Incenter $=\left(\frac{a·x_1+b·x_2+c·x_3}{a+b+c},\frac{a·y_1+b·y_2+c·y_3}{a+b+c}\right)$

Here, in the given triangle,

$$\begin{gathered} \left(x_1,y_1\right)=\left(0,0\right) \\ \left(x_2,y_2\right)=\left(5,12\right) \\ \left(x_3,y_3\right)=\left(16,12\right) \end{gathered}$$

And,

$$\begin{gathered} a=11 \\ b=20 \\ c=13 \end{gathered}$$

So, incenter of $△ABC$ $=\left(\frac{11\times 0+20\times 5+13\times 16}{11+20+13},\frac{11\times 0+20\times 12+13\times 12}{11+20+13}\right)$

$\Rightarrow$ incenter of $△ABC$ $=\left(\frac{0+100+208}{44},\frac{0+240+156}{44}\right)$

$\Rightarrow$ incenter of $△ABC$ $=\left(\frac{308}{44},\frac{396}{44}\right)$

$\Rightarrow$ incenter of $△ABC$ $=\left(7,9\right)$

Now, we know that,

Orthocenter of $△I_1{I}_2{I}_3=$ Incenter of $△ABC$

$\Rightarrow$Orthocenter of $△I_1{I}_2{I}_3=\left(7,9\right)$

Hence, the orthocenter of $△I_1{I}_2{I}_3$ is $\left(7,9\right)$.