A triangle has corners at (2 , 1 ), ( 5 , 6), and ( 8 , 5 ). What are the endpoints and lengths of the triangle's perpendicular bisectors?

Jan 17, 2016

Endpoints at pairs of coordinates [$\left(3.5 , 3.5\right) , \left(\frac{89}{19} , \frac{53}{19}\right)$],[$\left(6.5 , 5.5\right) , \left(\frac{41}{7} , \frac{25}{7}\right)$] and [$\left(5 , 3\right) , \left(\frac{77}{19} , \frac{84}{19}\right)$], and lengths equal to $9 \cdot \frac{\sqrt{34}}{38} , 9 \cdot \frac{\sqrt{10}}{14} \mathmr{and} 9 \cdot \frac{\sqrt{13}}{19}$.

Explanation:

Repeating the points
$A \left(2 , 1\right) , B \left(5 , 6\right) , C \left(8 , 5\right)$

Midpoints
${M}_{A B} \left(3.5 , 3.5\right)$, ${M}_{B C} \left(6.5 , 5.5\right)$, ${M}_{C A} \left(5 , 3\right)$

Slopes of segments ($k = \frac{\Delta y}{\Delta x}$, $p = - \frac{1}{k}$)
$A B \to {k}_{1} = \frac{6 - 1}{5 - 2} = \frac{5}{3} \to {p}_{1} = - \frac{3}{5}$
$B C \to {k}_{2} = \frac{5 - 6}{8 - 5} = - \frac{1}{3} \to {p}_{2} = 3$
$C A \to {k}_{3} = \frac{1 - 5}{2 - 8} = \frac{- 4}{- 6} = \frac{2}{3} \to {p}_{3} = - \frac{3}{2}$

Which of the two other sides does the perpendicular line to a side and bisecting it meet? The LONG side. Then we need to know the lengths of the sides of the triangle.
$A B = \sqrt{{\left(5 - 2\right)}^{2} + {\left(6 - 1\right)}^{2}} = \sqrt{9 + 25} = \sqrt{34} \cong 5.8$
$B C = \sqrt{{\left(8 - 5\right)}^{2} + {\left(5 - 6\right)}^{2}} = \sqrt{9 + 1} = \sqrt{10} \cong 3.1$
$C A = \sqrt{{\left(8 - 2\right)}^{2} + {\left(5 - 1\right)}^{2}} = \sqrt{36 + 16} = \sqrt{52} \cong 7.2$
=> $C A > A B > B C$

So
line 1 perpendicular to AB meets side CA
line 2 perpendicular to BC meets side CA
line 3 perpendicular to AC meets side AB

We need the equations of the lines in which the sides AB and CA lay and the equations of the 3 perpendicular lines

Equation of the line that supports side:
$A B \to \left(y - 1\right) = \left(\frac{5}{3}\right) \left(x - 2\right)$ => $y = \frac{5 x - 10}{3} + 1$ => $y = \frac{5 x - 7}{3}$ [a]
$C A \to \left(y - 4\right) = \left(\frac{2}{3}\right) \left(x - 2\right)$=>$y = \frac{2 x - 4}{3} + 1$=>$y = \frac{2 x - 1}{3}$[c]

Equation of the line (passing through midpoint) perpendicular to side:
$A B \to \left(y - 3.5\right) = \left(- \frac{3}{5}\right) \left(x - 3.5\right)$ => $y = \frac{- 3 x + 10.5}{5} + 3.5$ => $y = \frac{- 3 x + 28}{5}$ [1]
$B C \to \left(y - 5.5\right) = 3 \left(x - 6.5\right)$ => $y = 3 x - 19.5 + 5.5$ => $y = 3 x - 14$ [2]
$C A \to \left(y - 3\right) = - \left(\frac{3}{2}\right) \left(x - 5\right)$ => $y = \frac{- 3 x + 15}{2} + 3$ => $y = \frac{- 3 x + 21}{2}$ [3]

Finding the interceptions on sides AB and CA

Combining equations [1] and [c]

{y=(-3x+28)/5
{y=(2x-1)/3 => $\frac{- 3 x + 28}{5} = \frac{2 x - 1}{3}$ => $- 9 x + 84 = 10 x - 5$ => $19 x = 89$ => $x = \frac{89}{19}$
$\to y = \frac{2 \cdot \frac{89}{19} - 1}{3} = \frac{178 - 19}{57} = \frac{159}{57}$ => $y = \frac{53}{19}$

We've found $R \left(\frac{89}{19} , \frac{53}{19}\right)$
The distance between ${M}_{A B}$ and R is
$d 1 = \sqrt{{\left(\frac{89}{19} - \frac{7}{2}\right)}^{2} + {\left(\frac{53}{19} - \frac{7}{2}\right)}^{2}} = \frac{\sqrt{{\left(178 - 133\right)}^{2} + {\left(106 - 133\right)}^{2}}}{38} = \frac{\sqrt{2025 + 729}}{38} = \frac{\sqrt{2754}}{38} = 9 \cdot \frac{\sqrt{34}}{38} \cong 1.381$

Combining equations [2] and [c]

{y=3x-14
{y=(2x-1)/3 => $3 x - 14 = \frac{2 x - 1}{3}$ => $9 x - 42 = 2 x - 1$ => $7 x = 41$ => $x = \frac{41}{7}$
$\to y = 3 \cdot \frac{41}{7} - 14 = \frac{123 - 98}{7}$ => $y = \frac{25}{7}$

We've found $S \left(\frac{41}{7} , \frac{25}{7}\right)$
The distance between ${M}_{B C}$ and $S$ is:
$d 2 = \sqrt{{\left(\frac{41}{7} - \frac{13}{2}\right)}^{2} + {\left(\frac{25}{7} - \frac{11}{2}\right)}^{2}} = \frac{\sqrt{{\left(82 - 91\right)}^{2} + {\left(50 - 77\right)}^{2}}}{14} = \frac{\sqrt{81 + 729}}{14} = \frac{\sqrt{810}}{14} = 9 \cdot \frac{\sqrt{10}}{14} \cong 2.032$

Combining the equations [3] and [a]

{y=(-3x+21)/2
{y=(5x-7)/3#
=> $\frac{- 3 x + 21}{2} = \frac{5 x - 7}{3}$ => $- 9 x + 63 = 10 x - 14$ => $19 x = 77$ => $x = \frac{77}{19}$
$\to y = \frac{- 3 \cdot \frac{77}{19} + 21}{2} = \frac{- 231 + 399}{38} = \frac{168}{38}$ => $y = \frac{84}{19}$

We've found $T \left(\frac{77}{19} , \frac{84}{19}\right)$
The distance between ${M}_{C A}$ and T is
$d 3 = \sqrt{{\left(\frac{77}{19} - 5\right)}^{2} + {\left(\frac{84}{19} - 3\right)}^{2}} = \frac{\sqrt{{\left(77 - 95\right)}^{2} + {\left(84 - 57\right)}^{2}}}{19} = \frac{\sqrt{324 + 729}}{19} = \frac{\sqrt{1053}}{19} = 9 \cdot \frac{\sqrt{13}}{19} \cong 1 , 708$