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

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?

Caroline Aucoin · Accepted Answer

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#.

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# Thus, line 1 intersects with side CA, line 2 intersects with side BC, and line 3 intersects with side AB. We require the equations of the three perpendicular lines as well as the lines in which the sides AB and CA lie. 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] identifying the interceptions on the AB and CA sides Putting equations [1] and [c] together #{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# Putting equations [2] and [c] together #{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# Integrating the formulas [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#

James Atkins · Answer

undefined To find the endpoints of the perpendicular bisectors of the triangle, we first need to determine the midpoints of each side of the triangle. Then, we'll find the slopes of the sides and use those slopes to find the slopes of the perpendicular bisectors. Afterward, we'll use the midpoints and slopes to find the equations of the perpendicular bisectors. Finally, we'll solve the equations to find the endpoints of the bisectors.

Midpoint of the side joining (2, 1) and (5, 6):
Midpoint = ((2 + 5)/2, (1 + 6)/2) = (3.5, 3.5)

Midpoint of the side joining (5, 6) and (8, 5):
Midpoint = ((5 + 8)/2, (6 + 5)/2) = (6.5, 5.5)

Midpoint of the side joining (8, 5) and (2, 1):
Midpoint = ((8 + 2)/2, (5 + 1)/2) = (5, 3)

Slope of the side joining (2, 1) and (5, 6):
Slope = (6 - 1)/(5 - 2) = 5/3

Slope of the perpendicular bisector of this side = -1/(5/3) = -3/5

Equation of the perpendicular bisector passing through (3.5, 3.5) with slope -3/5:
y - 3.5 = (-3/5)(x - 3.5)

Slope of the side joining (5, 6) and (8, 5):
Slope = (5 - 6)/(8 - 5) = -1/3

Slope of the perpendicular bisector of this side = -1/(-1/3) = 3

Equation of the perpendicular bisector passing through (6.5, 5.5) with slope 3:
y - 5.5 = 3(x - 6.5)

Slope of the side joining (8, 5) and (2, 1):
Slope = (1 - 5)/(2 - 8) = 4/-6 = -2/3

Slope of the perpendicular bisector of this side = -1/(-2/3) = 3/2

Equation of the perpendicular bisector passing through (5, 3) with slope 3/2:
y - 3 = (3/2)(x - 5)

Solving these equations will give us the endpoints of the perpendicular bisectors.