Question

Find the coordinate of the circumcentre of a triangle whose vertices are $(-3,1)$, $(0,-2)$ and $(1,3)$.

Answer 1

Say $A,B,C$ are the vertices then

$A=(-3,1)$, $B=(0,-2)$, $C=(1,3)$

So, mid point of $AB=(\frac{-3+0}{2},\frac{1-2}{2})=(-1.5,-.5)$

Slope of $AB=\frac{-2-(-1.5)}{0-(-.5)}=-1$
Slope of the bisector is the negative reciprocal of the given slope.
So, the slope of the perpendicular bisector $=1$

Equation of $AB$ with slope $1$ and the coordinates $(-1.5,-0.5)$ is,
$[y–(-0.5\left)\right]=1[x–(-1.5\left)\right]$
$y-x=1$ ......(1)

Similarly, for $AC$
Mid point of $AC=(\frac{-3+1}{2},\frac{1+3}{2})=(-1,2)$

Slope of $AC=\left[\frac{3-(-1.5)}{1-(-.5)}\right]=3$
Slope of the bisector is the negative reciprocal of the given slope.
So, the slope of the perpendicular bisector $=-3$
Equation of $AC$ with slope $-3$ and the coordinates $(-1,2)$ is,
$(y–2)=-3(x+1)$
$y–2=-3x-3$
$y+3x=-1$ ......(2)

On subtracting equation (1) from (2), We get
$4x=-2orx=-0.5$

Substitute the value of $x$ in equation (1), we get
$y-(-0.5)=-1$
$y=1-0.5=0.5$

So, the circumcentre is $(-0.5,0.5)$.