Find the coordinate of the circumcenter of the triangle with vertices $A (2, 5)$ , $B (6, 6)$ and $C (12, 3)$ .

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Solution

Step-1: Define the circumcenter of $△ A B C$ :
The point of intersection perpendicular bisector of sides of $△ A B C$ is called the circumcenter of $△ A B C$ .
Step-2: To find the coordinate of circumcenter, find the equation of perpendicular bisector of side $A B$ :
Given, vertices of the triangle are $A (2, 5)$ , $B (6, 6)$ and $C (12, 3)$ .
Let $P (x_{1}, y_{1})$ be the point of coordinate of circumcenter.
And, $D$ be the mid-point of $A B$ :
$\begin{array}{rcl} D (x, y) & = & D (\frac{2 + 6}{2}, \frac{5 + 6}{2}) \\ = & D (4, \frac{11}{2}) \end{array}$
$∵ D P$ is perpendicular to $A B$ , the slope of $D P$ is negative reciprocal of $A B$ .
$\begin{array}{rcl} Slopeof D P & = & - \frac{1}{Slopeof A B} \\ = & - \frac{1}{\frac{6 - 5}{6 - 2}} ∵ Slopeof A B, m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}; (x_{1}, y_{1}) = (2, 5), (x_{2}, y_{2}) = (6, 6) \\ = & - 4 \end{array}$
The equation of $D P$ is defined as:
$\begin{array}{rcl} y - \frac{11}{2} & = & - 4 (x - 4) \\ \Rightarrow & 8 x + 2 y = 43 . . . (1) \end{array}$
Hence the equation of perpendicular bisector $D P$ is $8 x + 2 y = 43$ .
Step-3: Find the equation of perpendicular bisector of $A C$ :
Let $E P$ be the perpendicular bisector of $A C$ .
The coordinate of point $E$ is the mid point of $A C$ .
$\begin{array}{rcl} E (x, y) & = & E (\frac{2 + 12}{2}, \frac{5 + 3}{2}) \\ = & E (7, 4) \end{array}$
$∵ E P ⊥ A C$ , the slope of $E P$ is negative reciprocal of $A C$ .
$\begin{array}{rcl} Slopeof E P & = & - \frac{1}{Slopeof A C} \\ = & - \frac{1}{\frac{3 - 5}{12 - 2}} \\ = & 5 \end{array}$
The equation of $E P$ is defined as:
$\begin{array}{rcl} y - 4 & = & 5 (x - 7) \\ \Rightarrow & 5 x - y = 31 . . . (2) \end{array}$
Step-4: Find the solution of $(1)$ and $(2)$ :
Multiply $(2)$ by $2$ and add to $(1)$ :
$10 x - 2 y = 62 + 8 x + 2 y = 43 18 x = 105$
And,
$\begin{array}{rcl} \Rightarrow & x = \frac{105}{18} \\ \Rightarrow & x = \frac{35}{6} \end{array}$
Find $y$ by substituting $x = \frac{35}{6}$ in $(2)$ :
$\begin{array}{rcl} 5 (\frac{35}{6}) - y & = & 31 \\ \Rightarrow & y = \frac{11}{6} \end{array}$
The solution of $(1)$ and $(2)$ is $(\frac{35}{6}, \frac{11}{6})$ .
Hence, the required coordinate of circumcenter is $(\frac{35}{6}, \frac{11}{6})$ .

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