Question

Find the circumcentre of the triangle formed by the points $(2,3)$, $(1,-5)$ and $(-1,4)$

Answer 1

Step 1. Find the equation of the perpendicular bisector of the line $AB$

Given three points is $A(2,3)$, $B(1,-5)$ and $C(-1,4)$

Mid point of $AB,D=\left(\frac{2+1}{2},\frac{3-5}{2}\right)=\left(\frac{3}{2},-1\right)$

Slope of $AB=\frac{y_2-y_1}{x_2-x_1}=\frac{-5-3}{1-2}=\frac{-8}{-1}=8$

Therefore, slope of the perpendicular bisector $=-\frac{1}{8}$

Therefore equation of the perpendicular bisector of $AB$ with slope $=-\frac{1}{8}$ and passing coordinate $\left(\frac{3}{2},-1\right)$

$$\begin{gathered} y-y_1=m(x-x_1) \\ \Rightarrow y+1=-\frac{1}{8}(x-\frac{3}{2}) \\ \Rightarrow y+1=-\frac{1}{16}(2x-3) \\ \Rightarrow 16y+16=-2x+3 \\ \Rightarrow 2x+16y=-13\to \left(1\right) \end{gathered}$$

Step 2. Find the equation of the perpendicular bisector of the line $AC$

Mid point of $AC,E=\left(\frac{2-1}{2},\frac{3+4}{2}\right)=\left(\frac{1}{2},\frac{7}{2}\right)$

Slope of $AC=\frac{y_2-y_1}{x_2-x_1}=\frac{4-3}{-1-2}=\frac{-1}{3}$

Therefore, slope of the perpendicular bisector $=3$

Therefore equation of the perpendicular bisector of the line $AC$ with respect to slope $=3$ and passing through the coordinate $\left(\frac{1}{2},\frac{7}{2}\right)$

$$\begin{gathered} y-y_1=m(x-x_1) \\ \Rightarrow y-\frac{7}{2}=3(x-\frac{1}{2}) \\ \Rightarrow 2y-7=3(2x-1) \\ \Rightarrow 2y-7=6x-3 \\ \Rightarrow 6x-2y=-4\to \left(2\right) \end{gathered}$$

Step 3. Find the circumcenter

Now multiplying equation $\left(2\right)$ with $8$ and adding it with equation $\left(1\right)$ we get

$$\begin{gathered} 48x-16y=-32\to \left(3\right) \\ 2x+16y=-13 \end{gathered}$$

We get $x=-\frac{9}{10}$

Putting the value of $x$ in equation $\left(1\right)$ we get $y=-\frac{7}{10}$

Hence, the circumcenter of triangle is $\left(\frac{-9}{10},\frac{-7}{10}\right)$.