Question

$A=(2,2),B=(2,5)$ and $C(5,2)$ form a triangle. The circumcentre of $\Delta ABC$ is

• A. $(3,3)$
• B. $(2,2)$
• C. $(3.5,3.5)$
• D. $(2.5,2.5)$

Answer 1

Answer: (C)

$(3.5,3.5)$

Given the three vertices of $△$ABC

A(2,2),B(2,5) and C(5,2)

Mid point of $AB=(\frac{2+2}{2},\frac{2+5}{2})=(2,\frac{7}{2})$

$\therefore$ Slope of $AB=\frac{y_2-y_1}{x_2-x_1}=\frac{5-2}{2-2}=0$

$\therefore$ Slope of perpendicular bisector =0

$\therefore$ Equation of AB with respect to slope and coordinates

$y-y_1=m(x-x_1)y-\frac{7}{2}=0(x-2)\Rightarrow y=\frac{7}{2}⟶\left(1\right)$

Similarly for AC,A(2,2) A(2,2) and C(5,2)

Mid point of $AC=(\frac{2+5}{2},\frac{2+2}{2})=(\frac{7}{2},2)$

$\therefore$ Slope of $AC+\frac{y_2-y_1}{x_2-x_1}=\frac{2-2}{5-2}=0$

$\therefore$ Slope of perpendicular bisector =0

$\therefore$ Equation of AC with respect to slope and coordinates ,

$y{-y}_1=m(x-x_1)\Rightarrow y-2=0\left(x\frac{-7}{2}\right)\therefore y=2⟶\left(2\right)$

For BC, B(2,5) and C(5,2)

Mid point of $BC=(\frac{2+5}{2},\frac{5+2}{2})=(\frac{7}{2},\frac{7}{2})$

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

$\therefore$ Slope of perpendicular bisector =1

$\therefore$ Equation of BC with respect to slope and coordinates

$y-y_1=m(x-x_1)y\frac{-7}{2}=1\left(x\frac{-7}{2}\right)2y-7=2x-7\Rightarrow x=y⟶\left(3\right)$

By comparing equation (1) and (3)

$y=\frac{7}{2}\Rightarrow (x,y)=\left(\frac{7}{2}\right)$

$\therefore$ Circumcenter of $△$ABC$=(x,y)=(\frac{7}{2},\frac{7}{2})$

$\therefore (3.5,3.5)$