# A triangle has corners at (6 , 1 ), ( 4, 2 ), and ( 2, 8 ). What are the endpoints and lengths of the triangle's perpendicular bisectors?

Jan 22, 2016

Endpoints at [$\left(5 , 1.5\right) , \left(\frac{16}{3} , 7.5\right)$], [$\left(3 , 5\right) , \left(3.6 , 5.2\right)$] and [$\left(4 , 4.5\right) , \left(3.3 , 4.1\right)$], lengths measuring $\frac{5 \sqrt{13}}{3} , \frac{\sqrt{10}}{5}$ and $\frac{\sqrt{65}}{10}$

#### Explanation:

Repeating the points
$A \left(6 , 1\right) , B \left(4 , 2\right) , C \left(2 , 8\right)$

Midpoints
${M}_{A B} \left(5 , 1.5\right)$, ${M}_{B C} \left(3 , 5\right)$, ${M}_{C A} \left(4 , 4.5\right)$

Slopes of segments ($k = \frac{\Delta y}{\Delta x}$, $p = - \frac{1}{k}$)
$A B \to {k}_{1} = \frac{2 - 1}{4 - 6} = \frac{1}{- 2} = - \frac{1}{2} \to {p}_{1} = 2$
$B C \to {k}_{2} = \frac{8 - 2}{2 - 4} = \frac{6}{- 2} = - 3 \to {p}_{2} = \frac{1}{3}$
$C A \to {k}_{3} = \frac{1 - 8}{6 - 2} = - \frac{7}{4} \to {p}_{3} = \frac{4}{7}$

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.
$A B = \sqrt{{\left(4 - 6\right)}^{2} + {\left(2 - 1\right)}^{2}} = \sqrt{4 + 1} = \sqrt{5} \cong 2.2$
$B C = \sqrt{{\left(2 - 4\right)}^{2} + {\left(8 - 2\right)}^{2}} = \sqrt{4 + 36} = \sqrt{40} = 2 \sqrt{10} \cong 6.3$
$C A = \sqrt{{\left(6 - 2\right)}^{2} + {\left(1 - 8\right)}^{2}} = \sqrt{16 + 49} = \sqrt{65} \cong 8.1$
=> $C A > B C > A B$

So
line [1] perpendicular to AB meets side CA [c]
line [2] perpendicular to BC meets side CA [c]
line [3] perpendicular to AC meets side BC [b]

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

Equation of the line that supports side:
$B C \to \left(y - 2\right) = - 3 \left(x - 4\right)$ => $y = - 3 x + 12 + 2$ => $y = - 3 x + 14$ [a]
$C A \to \left(y - 1\right) = - \left(\frac{7}{4}\right) \left(x - 6\right)$=>$y = \frac{- 7 x + 42}{4} + 1$=>$y = \frac{- 7 x + 46}{4}$[c]

Equation of the line (passing through midpoint) perpendicular to side:
$A B \to \left(y - 1.5\right) = 2 \left(x - 5\right)$ => $y = 2 x - 10 + 1.5$ => $y = 2 x - 8.5$ [1]
$B C \to y - 5 = \frac{1}{3} \left(x - 3\right)$ => $y = \frac{x - 3}{3} + 5$ => $y = \frac{x + 12}{3}$ [2]
$C A \to \left(y - 4.5\right) = \left(\frac{4}{7}\right) \left(x - 4\right)$ => $y = \frac{4 x - 16}{7} + 4.5$ => $y = \frac{4 x + 15.5}{7}$ [3]

Finding the interceptions on sides BC and CA

Combining equations [1] and [c]

{y=2x-8.5
{y=(-7x+46)/4 => $2 x - 8.5 = \frac{- 7 x + 46}{4}$ => $8 x + 34 = - 7 x + 46$ => $15 x = 80$ => $x = \frac{16}{3}$
$\to y = 2 \cdot \frac{16}{3} - \frac{17}{2} = \frac{96 - 51}{6} = \frac{45}{6}$ => $y = \frac{15}{2}$

We've found $R \left(\frac{16}{3} , 7.5\right)$
The distance between ${M}_{A B}$ and R is
$d 1 = \sqrt{{\left(\frac{16}{3} - 5\right)}^{2} + {\left(\frac{15}{2} - \frac{3}{2}\right)}^{2}} = \sqrt{{\left(\frac{1}{3}\right)}^{2} + {6}^{2}} = \frac{\sqrt{1 + 324}}{3} = \frac{\sqrt{325}}{3} = \frac{5 \cdot \sqrt{13}}{3} \cong 6.009$

Combining equations [2] and [c]

{y=(x+12)/3
{y=(-7x+46)/4 => $\frac{x + 12}{3} = \frac{- 7 x + 46}{4}$ => $4 x + 48 = - 21 x + 138$ => $25 x = 90$ => $x = \frac{18}{5}$
$\to y = \frac{\frac{18}{5} + 12}{3} = \frac{18 + 60}{15}$ => $y = \frac{26}{5}$

We've found $S \left(3.6 , 5.2\right)$
The distance between ${M}_{B C}$ and $S$ is:
$d 2 = \sqrt{{\left(\frac{18}{5} - 3\right)}^{2} + {\left(\frac{26}{5} - 5\right)}^{2}} = \frac{\sqrt{{3}^{2} + {1}^{2}}}{5} = \frac{\sqrt{10}}{5} \cong .632$

Combining the equations [3] and [a]

{y=(4x+15.5)/7
{y=-3x+14# => $\frac{4 x + 15.5}{7} = - 3 x + 14$ => $4 x + 15.5 = - 21 + 98$ => $25 = 82 , 5$ => $x = 3.3$
$\to y = - 3 \cdot 3.3 + 14$ => $y = 4.1$

We've found $T \left(3.3 , 4.1\right)$
The distance between ${M}_{C A}$ and T is
$d 3 = \sqrt{{\left(3.3 - 4\right)}^{2} + {\left(4.1 - 4.5\right)}^{2}} = \sqrt{{.7}^{2} + {.4}^{2}} = \sqrt{.49 + .16} = \sqrt{.65} = \frac{\sqrt{65}}{10} \cong .806$