# A triangle has corners at (2 , 4 ), ( 3, 1 ), and ( 8, 3 ). What are the endpoints and lengths of the triangle's perpendicular bisectors?

Jun 20, 2016

One end-pt. of all perp. bsctrs. is $O \left(\frac{169}{34} , \frac{113}{34}\right)$ & other end. pts. are D(11/2,2),E(5,7/2), & F(5/2,5/2).

Length of one perp.bsctr. $O D$ is $\frac{\sqrt{2349}}{34.}$

#### Explanation:

Let us name the vertices of $\Delta$ as $A \left(2 , 4\right) , B \left(3 , 1\right) , C \left(8 , 3\right)$ & let the mid-pts. of sides $B C , C A , A B$ be $D , E , F$ resp.

Clearly, the mid-pts. are D(11/2,2),E(5,7/2), & F(5/2,5/2).

We know that three perp. bsctrs. of sides of a $\Delta$ are concurrent at a pt., known as the Circumcentre of $\Delta A B C .$ Let us call it $O .$

To find $O$, we find the eqns. of two perp.bstrs., namely, OD & OE.#

Eqn. of $O D$:-

$O D$ is perp. to $B C$, & slope of $B C$ is $\frac{3 - 1}{8 - 3} = \frac{2}{5} ,$ so, slope of $O D$ must be $- \frac{5}{2}$. In addition, $D \in O D .$

$\therefore$ eqn. of $O D$ is, $y - 2 = - \frac{5}{2} \left(x - \frac{11}{2}\right) ,$ or, $4 y - 8 = - 5 \left(2 x - 11\right) = - 10 x + 55 ,$ i.e., $10 x + 4 y = 63. \ldots \ldots \ldots . \left(1\right) .$

On the same line, we can work out the Eqn. of $O E$ as $12 x - 2 y = 53. \ldots . . \left(2\right)$

Solving (1) & (2), we get $O \left(\frac{169}{34} , \frac{113}{34}\right)$

Length $O D = \sqrt{{\left(\frac{169}{34} - \frac{11}{2}\right)}^{2} + {\left(\frac{113}{34} - 2\right)}^{2}} = \frac{\sqrt{2349}}{34}$