Answer to Question #139730 in Analytic Geometry for Aravind

Question

5)The coordinates of the vertices of a triangle are (3,2), (9,2), (6,5).
a)Find the equations of the perpendicular bisectors of triangle.
b)Find the co ordinates of the circumcentre

Answer 1

The coordinates of the vertices of a triangle are $A(3,2), B(9,2), C(6,5).$

a)

The midpoint of $AB$ is $M_1\big(\frac{3+9}{2},\ \frac{2+2}{2}\big)=M_1(6,\ 2)$

The slope of $AB$ is $\frac{2-2}{9-3}=\frac{0}{6}=0$. It means that $AB$ is a horizontal line and perpendicular bisector is a vertical line. The equation of the perpendicular bisector is $x=6$ .

The midpoint of $BC$ is $M_2\big(\frac{9+6}{2},\ \frac{2+5}{2}\big)=M_2(7.5,\ 3.5)$ 

The slope of $BC$ is $\frac{2-5}{9-6}=\frac{-3}{3}=-1$. The slope of the perpendicular bisector is $m_2=-\frac{1}{-1}=1$ .

The equation of the perpendicular bisector is $y-3.5=1\times (x-7.5)$ or $y=x-4$ .

The midpoint of $AC$ is $M_3\big(\frac{3+6}{2},\ \frac{2+5}{2}\big)=M_3(4,5,\ 3.5)$ 

The slope of $AC$ is $\frac{2-5}{3-6}=\frac{-3}{-3}=1$. The slope of the perpendicular bisector is $m_3=-\frac{1}{1}=-1$ .The equation of the perpendicular bisector is $y-3.5=-1\times (x-4.5)$ or $y=-x+8$ .

b)

The circumcenter is the point of concurrency of perpendicular bisectors of a triangle. We need to solve system of any two bisector equations to find the coordinates of the circumcenter:

$\begin{cases} x=6 \\ y=x-4 \end{cases}$ $\begin{cases} x=6 \\ y=2 \end{cases}$

The circumcenter is $O(6,\ 2)$ .

Answers: a) $x=6$ , $y=x-4$ , $y=-x+8$ ; b) $(6,\ 2)$ .