Question

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

Answer 1

Endpoints at pairs of coordinates [`(3.5,3.5), (89/19,53/19)`],[`(6.5,5.5),(41/7,25/7)`] and [`(5,3),(77/19,84/19)`], and lengths equal to `9*sqrt(34)/38, 9*sqrt(10)/14 and 9*sqrt(13)/19`.

Explanation:

Repeating the points
`A(2,1), B(5,6), C(8,5)`

Midpoints
`M_(AB) (3.5,3.5)`, `M_(BC) (6.5,5.5)`, `M_(CA) (5,3)`

Slopes of segments (`k=(Delta y)/(Delta x)`, `p=-1/k`)
`AB -> k_1=(6-1)/(5-2)=5/3 -> p_1=-3/5`
`BC -> k_2=(5-6)/(8-5)=-1/3 -> p_2=3`
`CA -> k_3=(1-5)/(2-8)=(-4)/(-6)=2/3 -> p_3=-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.
`AB=sqrt((5-2)^2+(6-1)^2)=sqrt (9+25)=sqrt(34)~=5.8`
`BC=sqrt((8-5)^2+(5-6)^2)=sqrt(9+1)=sqrt(10)~=3.1`
`CA=sqrt((8-2)^2+(5-1)^2)=sqrt(36+16)=sqrt(52)~=7.2`
=> `CA>AB>BC`

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:
`AB -> (y-1)=(5/3)(x-2)` => `y=(5x-10)/3+1` => `y=(5x-7)/3` [a]
`CA->(y-4)=(2/3)(x-2)`=>`y=(2x-4)/3+1`=>`y=(2x-1)/3`[c]

Equation of the line (passing through midpoint) perpendicular to side:
`AB -> (y-3.5)=(-3/5)(x-3.5)` => `y=(-3x+10.5)/5+3.5` => `y=(-3x+28)/5` [1]
`BC -> (y-5.5)=3(x-6.5)` => `y=3x-19.5+5.5` => `y=3x-14` [2]
`CA -> (y-3)=-(3/2)(x-5)` => `y=(-3x+15)/2+3` => `y=(-3x+21)/2` [3]

Finding the interceptions on sides AB and CA

Combining equations [1] and [c]

`{y=(-3x+28)/5`
`{y=(2x-1)/3` => `(-3x+28)/5=(2x-1)/3` => `-9x+84=10x-5` => `19x=89` => `x=89/19`
`-> y=(2*89/19-1)/3=(178-19)/57=159/57` => `y=53/19`

We've found `R(89/19,53/19)`
The distance between `M_(AB)` and R is
`d1=sqrt((89/19-7/2)^2+(53/19-7/2)^2)=sqrt((178-133)^2+(106-133)^2)/38=sqrt(2025+729)/38=sqrt(2754)/38=9*sqrt(34)/38~=1.381`

Combining equations [2] and [c]

`{y=3x-14`
`{y=(2x-1)/3` => `3x-14=(2x-1)/3` => `9x-42=2x-1` => `7x=41` => `x=41/7`
`-> y=3*41/7-14=(123-98)/7` => `y=25/7`

We've found `S(41/7,25/7)`
The distance between `M_(BC)` and `S` is:
`d2=sqrt((41/7-13/2)^2+(25/7-11/2)^2)=sqrt((82-91)^2+(50-77)^2)/14=sqrt(81+729)/14=sqrt(810)/14=9*sqrt(10)/14~=2.032`

Combining the equations [3] and [a]

`{y=(-3x+21)/2`
`{y=(5x-7)/3`
=> `(-3x+21)/2=(5x-7)/3` => `-9x+63=10x-14` => `19x=77` => `x=77/19`
`-> y=(-3*(77)/19+21)/2=(-231+399)/38=168/38` => `y=84/19`

We've found `T(77/19,84/19)`
The distance between `M_(CA)` and T is
`d3=sqrt((77/19-5)^2+(84/19-3)^2)=sqrt((77-95)^2+(84-57)^2)/19=sqrt(324+729)/19=sqrt(1053)/19=9*sqrt(13)/19~=1,708`