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

1 Answer
Jan 17, 2016

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#