Question

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

Answer 1

Endpoints at [`(5,1.5),(16/3,7.5)`], [`(3,5),(3.6,5.2)`] and [`(4,4.5),(3.3,4.1)`], lengths measuring `(5sqrt(13))/3,sqrt(10)/5` and `sqrt(65)/10`

Explanation:

Repeating the points
`A(6,1), B(4,2), C(2,8)`

Midpoints
`M_(AB) (5,1.5)`, `M_(BC) (3,5)`, `M_(CA) (4,4.5)`

Slopes of segments (`k=(Delta y)/(Delta x)`, `p=-1/k`)
`AB -> k_1=(2-1)/(4-6)=1/(-2)=-1/2 -> p_1=2`
`BC -> k_2=(8-2)/(2-4)=6/(-2)=-3 -> p_2=1/3`
`CA -> k_3=(1-8)/(6-2)=-7/4 -> p_3=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.
`AB=sqrt((4-6)^2+(2-1)^2)=sqrt (4+1)=sqrt(5)~=2.2`
`BC=sqrt((2-4)^2+(8-2)^2)=sqrt(4+36)=sqrt(40)=2sqrt(10)~=6.3`
`CA=sqrt((6-2)^2+(1-8)^2)=sqrt(16+49)=sqrt(65)~=8.1`
=> `CA>BC>AB`

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:
`BC -> (y-2)=-3(x-4)` => `y=-3x+12+2` => `y=-3x+14` [a]
`CA-> (y-1)=-(7/4)(x-6)`=>`y=(-7x+42)/4+1`=>`y=(-7x+46)/4`[c]

Equation of the line (passing through midpoint) perpendicular to side:
`AB -> (y-1.5)=2(x-5)` => `y=2x-10+1.5` => `y=2x-8.5` [1]
`BC ->y-5=1/3(x-3)` => `y=(x-3)/3+5` => `y=(x+12)/3` [2]
`CA -> (y-4.5)=(4/7)(x-4)` => `y=(4x-16)/7+4.5` => `y=(4x+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` => `2x-8.5=(-7x+46)/4` => `8x+34=-7x+46` => `15x=80` => `x=16/3`
`-> y=2*16/3-17/2=(96-51)/6=45/6` => `y=15/2`

We've found `R(16/3,7.5)`
The distance between `M_(AB)` and R is
`d1=sqrt((16/3-5)^2+(15/2-3/2)^2)=sqrt((1/3)^2+6^2)=sqrt(1+324)/3=sqrt(325)/3=(5*sqrt(13))/3~=6.009`

Combining equations [2] and [c]

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

We've found `S(3.6,5.2)`
The distance between `M_(BC)` and `S` is:
`d2=sqrt((18/5-3)^2+(26/5-5)^2)=sqrt(3^2+1^2)/5=sqrt(10)/5~=.632`

Combining the equations [3] and [a]

`{y=(4x+15.5)/7`
`{y=-3x+14` => `(4x+15.5)/7=-3x+14` => `4x+15.5=-21+98` => `25=82,5` => `x=3.3`
`-> y=-3*3.3+14` => `y=4.1`

We've found `T(3.3,4.1)`
The distance between `M_(CA)` and T is
`d3=sqrt((3.3-4)^2+(4.1-4.5)^2)=sqrt(.7^2+.4^2)=sqrt(.49+.16)=sqrt(.65)=sqrt(65)/10~=.806`