A triangle has corners at #(2 , 2 )#, ( 5, 6 )#, and #( 1, 4 )#. What are the endpoints and lengths of the triangle's perpendicular bisectors?

Question

A triangle has corners at #(2   , 2  )#, (  5,  6 )#, and #( 1,  4  )#.  What are the endpoints and lengths of the triangle's perpendicular bisectors?

Evelyn Arboleda · Accepted Answer

Each of the 3 segments have one endpoint at the midpoint of the given points and the other endpoint at #(2.5,4.75), (3.5,4) or (53/22,28/11)#, and lengths equal to #1.25, .5*sqrt(5) or 5*sqrt(5)/11#.

Repeating the points #A(2,2), B(5,6), C(1,4)# Midpoints # M_(AB) (3.5,4)#, #M_(BC) (3,5)#, #M_(CA) (1.5,3) # Slopes of segments (#k=(Delta y)/(Delta x)#, #p=-1/k#) #AB -> k_1=(6-2)/(5-2)=4/3 -> p_1=-3/4# #BC -> k_2=(4-6)/(1-5)=(-2)/-4=1/2 -> p_2=-2# #CA -> k_3=(4-2)/(1-2)=4/(-1)=-2 -> p_3=-1/2# Now to simplify the work that remains to do, be noted that the line perpendicular to a side meets one of the other sides (or both at the same time when it intercepts a vertex of the triangle). When it meets just one of the two other sides, which side is this? Answer: the LONG side (although that is intuitive it can be proved). Then we need to know the lengths of the sides of the triangle. #AB=sqrt((5-2)^2+(6-2)^2)=sqrt (9+16)=5# #BC=sqrt((1-5)^2+(4-5)^2)=sqrt(16+4)=2*sqrt(5)~=4.472# #CA=sqrt((1-2)^2+(4-2)^2)=sqrt(1+4)=sqrt(5)~=2.236# => #AB>BC>CA# Thus, line 1 intersects with BC, line 2 intersects with BC, and line 3 intersects with AB. Line 1 is perpendicular to AB. We require the equations of the three perpendicular lines as well as the lines in which the sides AB and BC lay. Equation of the line that supports side: #AB -> (y-2)=(4/3)(x-2)# => #y=(4x-8)/3+2# => #y=(4x-2)/3# [a] #BC->y-4=(1/2)(x-1)#=>#y=(x-1)/2+4#=>#y=(x+7)/2#[b] Equation of the line (passing through midpoint) perpendicular to side: #AB -> (y-4)=(-3/4)(x-3.5)# => #y=(-3x+10.5)/4+4# => #y=(-3x+26.5)/4# [1] #BC -> (y-5)=-2(x-3)# => #y=-2x+6+5# => #y=-2x+11# [2] #AC -> (y-3)=-(1/2)(x-1.5)# => #y=(-x+1.5)/2+3# => #y=(-x+7.5)/2# [3] identifying the interceptions on AB and BC sides Putting equations [b] and [1] together #{y=(x+7)/2# #{y=(-3x+26.5)/4# => #(x+7)/2=(-3x+26.5)/4# => #2x+14=-3x+26.5# => #5x=12.5# => #x=2.5# #-> y=(2.5+7)/2# => #y=4.75# We've found #R(2.5,4.75)# The distance between #M_(AB)# and R is #d1=sqrt((2.5-3.5)^2+(4.75-4)^2)=sqrt(1+.5625)=1.25# Putting equations [a] and [2] together #{y=(4x-2)/3# #{y=-2x+11# => #(4x-2)/3=-2x+11# => #4x-2=-6x+33# => #10x=35# => #x=3.5# #-> y=-2*3.5+11# => #y=4# We've found # S(3.5,4)# We can find the distance between #M_(BC)# and #S#: #d2=sqrt((3.5-3)^2+(4-5)^2)=sqrt(1.25)=5*sqrt(5)# Putting the equations [a] and [3] together #{y=(4x-2)/3# #{y=(-x+7.5)/2# => #(4x-2)/3=(-x+7.5)/2# => #8x-4=-3x+22.5# => #11x=26.5# => #x=53/22# #-> y=(-53/22+15/2)/2=(-53+165)/44# => #y=28/11# We've found #T(53/22, 28/11)# The distance between #M_(CA)# and T is #d3=sqrt((53/22-3/2)^2+(28/11-3)^2)=sqrt((10/11)^2+(5/11)^2)=sqrt(125)/11=5*sqrt(5)/11#

Caroline Aucoin · Answer

undefined To find the perpendicular bisectors of the triangle, we need to find the midpoint of each side and then determine the slope of each side. After that, we can find the perpendicular bisectors, which will have slopes negative reciprocals of the slopes of the original sides. Finally, we find the equations of the lines passing through the midpoints with these slopes.

1. Midpoint of side 1 (between points (2, 2) and (5, 6)):
   Midpoint $ = \left(\frac{2 + 5}{2}, \frac{2 + 6}{2}\right) = (3.5, 4) $

2. Midpoint of side 2 (between points (2, 2) and (1, 4)):
   Midpoint $ = \left(\frac{2 + 1}{2}, \frac{2 + 4}{2}\right) = (1.5, 3) $

3. Midpoint of side 3 (between points (5, 6) and (1, 4)):
   Midpoint $ = \left(\frac{5 + 1}{2}, \frac{6 + 4}{2}\right) = (3, 5) $

Now, let's find the slopes of the sides:

1. Slope of side 1: $ m_1 = \frac{6 - 2}{5 - 2} = \frac{4}{3} $
2. Slope of side 2: $ m_2 = \frac{4 - 2}{1 - 2} = -2 $
3. Slope of side 3: $ m_3 = \frac{4 - 6}{1 - 5} = \frac{-2}{-4} = \frac{1}{2} $

Now, the slopes of the perpendicular bisectors will be negative reciprocals of these slopes:

1. Perpendicular bisector 1: Slope $ = -\frac{1}{\frac{4}{3}} = -\frac{3}{4} $
2. Perpendicular bisector 2: Slope $ = -\frac{1}{-2} = \frac{1}{2} $
3. Perpendicular bisector 3: Slope $ = -\frac{1}{\frac{1}{2}} = -2 $

Finally, we use the point-slope form to find the equations of the perpendicular bisectors:

1. Perpendicular bisector 1:
   $ y - 4 = -\frac{3}{4}(x - 3.5) $

2. Perpendicular bisector 2:
   $ y - 3 = \frac{1}{2}(x - 1.5) $

3. Perpendicular bisector 3:
   $ y - 5 = -2(x - 3) $

These are the equations of the perpendicular bisectors. You can solve them to find their endpoints and lengths.