Chapter 14.3, Problem 15PSC

To determine

To find the circumcenter of the triangle.

Answer to Problem 15PSC

  $\left(\frac{28}{5},\frac{8}{5}\right)$

Explanation of Solution

Given:

The triangle is ABC with $A=(1,3),B=(7,-3)$ and $C=(9,5)$

Calculation:

The circumcenter $C(x,y)$ of the triangle ABCis thepoint of intersection of all the three perpendicular bisectors of the sides of triangle.

  $\begin{array}{l}{D}_1=D_2=D_3\\ D_1=D_2\\ \sqrt{{(x-x_1)}^2+{(y-y_1)}^2}=\sqrt{{(x-x_2)}^2+{(y-y_2)}^2}\\ \Rightarrow \sqrt{{(x-1)}^2+{(y-3)}^2}=\sqrt{{(x-7)}^2+{(y+3)}^2}\\ $ \Rightarrow {(x-1)}^2+{(y-3)}^2={(x-7)}^2+{(y+3)}^2\mathrm{........}(1)$D_2=D_3\\ \sqrt{{(x-x_2)}^2+{(y-y_2)}^2}=\sqrt{{(x-x_3)}^2+{(y-y_3)}^2}\\ \sqrt{{(x-7)}^2+{(y+3)}^2}=\sqrt{{(x+3)}^2+{(y-5)}^2}\\ $ \Rightarrow {(x-7)}^2+{(y+3)}^2={(x+3)}^2+{(y-5)}^2\mathrm{..............}(2)$\end{array}$

Solving equation (1) and (2),-

  $\begin{array}{l}x-4=-4x+24\\ 5x=28\\ x=\frac{28}{5}\\ y=\frac{28}{5}-4=\frac{8}{5}\end{array}$

Therefore, the circumcenter is $\left(\frac{28}{5},\frac{8}{5}\right)$