# A triangle has corners at (5 , 3 ), ( 7 ,8), and ( 2, 1 )#. What are the endpoints and lengths of the triangle's perpendicular bisectors?

Apr 23, 2018

I get
$D = \left(\frac{9}{2} , \frac{9}{2}\right)$, $G = \left(\frac{241}{45} , \frac{35}{9}\right)$, $D G = \frac{11 \sqrt{74}}{90}$
$E = \left(\frac{7}{2} , 2\right)$, $I = \left(\frac{181}{58} , \frac{149}{58}\right)$, $E I = \frac{11 \sqrt{13}}{58}$
$F = \left(6 , \frac{11}{2}\right)$, $H = \left(\frac{97}{18} , \frac{517}{90}\right)$, $F H = 11 \frac{\sqrt{29}}{90}$

#### Explanation:

These are so hard -- who keeps posting them? Let's go with a figure this time. $A \left(5 , 3\right) , B \left(7 , 8\right) , C \left(2 , 1\right)$ First, the easy endpoints are the midpoints of the sides:

$D = \frac{B + C}{2} = \left(\frac{9}{2} , \frac{9}{2}\right)$
$E = \frac{A + C}{2} = \left(\frac{7}{2} , 2\right)$
$F = \frac{A + B}{2} = \left(6 , \frac{11}{2}\right)$

Let's write the equations for the sides:

$B C : \left(y - 1\right) \left(7 - 2\right) = \left(x - 2\right) \left(8 - 1\right) \quad \quad - 7 x + 5 y = - 9$
$A C : \left(y - 1\right) \left(5 - 2\right) = \left(x - 2\right) \left(3 - 1\right) \quad \quad - 2 x + 3 y = - 1$
$A B : \left(y - 3\right) \left(7 - 5\right) = \left(x - 5\right) \left(8 - 3\right) \quad \quad - 5 x + 2 y = - 19$

For perpendicular bisectors we swap the coefficients on $x$ and $y$ and negate one. We get the constant by plugging in the midpoint.

$D G \bot B C : 5 x + 7 y = 5 \left(\frac{9}{2}\right) + 7 \left(\frac{9}{2}\right) = 54$
$E I \bot A C : 3 x + 2 y = 3 \left(\frac{7}{2}\right) + 2 \left(2\right) = \frac{29}{2}$
$F H \bot A B : 2 x + 5 y = 2 \left(6\right) + 5 \left(\frac{11}{2}\right) = \frac{79}{2}$

$G$ is the meet of $D G$ and $A B$:
$5 x + 7 y = 54$
$- 5 x + 2 y = - 19$
$9 y = 35$
$y = \frac{35}{9}$
$x = \frac{1}{5} \left(54 - 7 \left(\frac{35}{9}\right)\right) = \frac{241}{45}$

$G = \left(\frac{241}{45} , \frac{35}{9}\right)$

$D G = \sqrt{{\left(\frac{9}{2} - \frac{241}{45}\right)}^{2} + {\left(\frac{9}{2} - \frac{35}{9}\right)}^{2}} = \frac{11 \sqrt{74}}{90}$

Phew. That's one, though it may not be right.

$I$ is the meet of $E I$ with $B C$
$3 x + 2 y = \frac{29}{2}$
$- 7 x + 5 y = - 9$
$30 x + 20 y = 145$
$- 28 x + 20 y = - 36$
$58 x = 181$
$x = \frac{181}{58}$
$y = \frac{149}{58}$

$I = \left(\frac{181}{58} , \frac{149}{58}\right)$

$E I = \sqrt{{\left(\frac{7}{2} - \frac{181}{58}\right)}^{2} + {\left(2 - \frac{149}{58}\right)}^{2}} = \frac{11 \sqrt{13}}{58}$

$F H$ meets $B C$ at $H$
$2 x + 5 y = \frac{79}{2}$
$- 7 x + 5 y = - 9$
$9 x = \frac{97}{2}$
$x = \frac{97}{18}$
$y = \frac{517}{90}$

$H = \left(\frac{97}{18} , \frac{517}{90}\right)$

$F H = \sqrt{{\left(6 - \frac{97}{18}\right)}^{2} + {\left(\frac{11}{2} - \frac{517}{90}\right)}^{2}} = 11 \frac{\sqrt{29}}{90}$