Question

Consider the triangle $ABC$ with vertices $A(-2,3),B(2,1)$ and $C(1,2)$.

What is the circumcentre of the triangle $ABC$?

• A. $(-2,-2)$
• B. $(2,2)$
• C. $(-2,2)$
• D. $(2,-2)$

Answer 1

The correct option is A $(-2,-2)$

Given points are,
$A=(-2,3),B=(2,1),C=(1,2)$
To find out the circumcenter we have to solve any two bisector equations and find out the intersection points.

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

Slope of $AB$ is $\frac{1-3}{2+2}=-\frac{1}{2}$

Slope of the bisector is the negative reciprocal of the given slope.
So, the slope of the perpendicular bisector $=2$

$\Rightarrow$ Equation of a line(perpendicular to AB) with slope $2$ and the coordinates $(0,2)$ is,

$y-2=2x$

$\Rightarrow 2x-y+2=0$ ........ $\left(i\right)$

Similarly,

So, mid-point of BC = $(\frac{1+2}{2},\frac{2+1}{2})=(3/2,3/2)$

Slope of $BC$ is $\frac{2-1}{1-2}=-1$

So, the slope of the perpendicular bisector $=1$

Equation of a line(perpendicular to BC) with slope $1$ and the coordinates $(3/2,3/2)$ is,

$(y-3/2)=(x-3/2)$

$\Rightarrow x=y$ ........ $\left(ii\right)$

Solving Eq $\left(i\right)$ and $\left(ii\right)$ gives $(x,y)=(-2,-2)$ which is the circumcentre of the given triangle.