Question

The circumcentre of a triangle , whose vertices are $(3,0),(-1,-6)$ and $(4,-1)$ is

• A. $(1,3)$
• B. $(3,1)$
• C. $(-1,3)$
• D. $(1,-3)$

Answer 1

The correct option is D $(1,-3)$

Given,

Cordinates of vertices of triangle are $A(3,0),B(-1,-6)andC(4,-1)$.

Distance AB=$\sqrt{(3-(-1){)}^2+(0+6{)}^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$

Distance BC=$\sqrt{(-1-4{)}^2+(-6+1{)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$

Distance AC=$\sqrt{(3-4{)}^2+(0+1{)}^2}=\sqrt{1+1}=\sqrt{2}$

Here,

$B{C}^2+A{C}^2=(\sqrt{2}{)}^2+(5\sqrt{2}{)}^2=50+2=52=A{B}^2$

So $\Delta ABC$ is right angle triangle with $\angle C={90}^{\circ }$

Therefore Circumcenter lies at the middle of the hypotenuse AB

Coordinates of circumcenter=$(\frac{3-1}{2},\frac{0-6}{2})=(\frac{2}{2},\frac{-6}{2})=(1,-3)$