Question

Find the circumcentre of the triangle whose vertices are $(2,1),(5,2)$ and $(3,4)$.

Answer 1

$A(2,1),B(5,2),C(3,4)$

Mid point of $AB=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{7}{2},\frac{3}{2})$

Slope of $AB=\left(\frac{y_1-y_2}{x_2-x_1}\right)=\frac{1}{3}$

$(y-y_1)=m(x-x_1)$

$(y-\frac{3}{2})=(-3)(x-\frac{7}{2})$

$\to 3x+y-12=0$...(1)

Mid point of $AC=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{5}{2},\frac{5}{2})$

Slope of $AC=\left(\frac{y_1-y_2}{x_2-x_1}\right)=3$

$(y-y_1)=m(x-x_1)$

$(y-\frac{5}{2})=\frac{-1}{3}(x-\frac{5}{2})$

$\to x+3y-10=0$...(2)

Solving (1) and (2), we get,

$y=\frac{9}{4},x=\frac{13}{4}$

$\therefore (\frac{13}{4},\frac{9}{4})$ is the circumcenter.