A triangle has corners at #(2 , 8 )#, ( 2, 3 )#, and #( 5, 4 )#. What are the endpoints and lengths of the triangle's perpendicular bisectors?

1 Answer
Rui D.
Jan 13, 2016

Each of the 3 segments have one endpoint at the midpoint of the given points and the other endpoint at #(3.875,5.5), (2,8) or (2,4.875)#, and lengths equal to #3.875, 1.5*sqrt(10) or 1.875#.

Explanation:

Repeating the points
#A(2,8), B(2,3), C(5,4)#

Midpoints
# M_(AB) (2,5.5)#, #M_(BC) (3.5,3.5)#, #M_(CA) (3.5,6) #

Slopes of segments (#k=(Delta y)/(Delta x)#, #p=-1/k#)
#AB -> "parallel to axis"-Y -> x=Const -> p_1=0#
#BC -> k_2=(4-3)/(5-2)=1/3 -> p_2=-3#
#CA -> k_3=(8-4)/(2-5)=-4/3 -> p_3=3/4#

(The following lines are: either (when marked with a number) supporting the triangle's sides AB, BC or CA or (when marked with a letter) perpendicular to one of these sides.)

Equations of lines through #M_(AB)(2,5.5)#
#x=2# [1]
#y=5.5# [a]

Equations of lines through #M_(BC)(3.5,3.5)#, using #k_2 and p_2#
#(y-3)=(1/3)(x-2)# => #y=(x/3)-2/3+3# => #y=(x+7)/3# [2]

#(y-3.5)=-3(x-3.5)# => #y=-3x+10.5+3.5# => #y=-3x+14# [b]

Equations of lines through #M_(CA) (3.5,6)#, using #k_3 and p_3#
#(y-8)=(-4/3)(x-2)# => #y=-(4x)/3+8/3+8# => #y=-(4x)/3+32/3# [3]

#(y-6)=(3/4)(x-3.5)# => #y=(3x)/4-2.625+6# => #y=(3x)/4+3.375# [c]

In a triangle, a line perpendicular to a side meets the two lines that support the other sides, but it meets one of these two lines inside or outside the triangle (when it meets the other line outside the triangle, the point of interception isn't valid for this kind of problem).

Finding the interceptions of a line (perpendicular to a side) with the lines supporting the two other sides.

Combining equations [a] and [2]

#{y=5.5#
#{y=(x+7)/3# => #5.5=(x+7)/3# => #16.5=x+7# => #x=9.5# (NOT VALID: out of the range of points B to C]

Combining equations [a] and [3]

#{y=5.5#
#{y=-(4x)/3+32/3# => #5.5=(-4x+32)/3# => #16.5=-4x+32# => #4x=15.5# => #x=3.875#

We've found # R(3.875, 5.5)#
We can find the distance between #M_(AB)# and #R#:
#d1=sqrt((3.875-2)^2+(5.5-5.5)^2=3.875#

Combining equations [b] and [1]

#{y=-3x+14#
#{x=2#
=> #y=-6+14# => #y=8# (this is in the vertex A, but it's OK because it's still in the triangle)

We've found #S(2,8)#
The distance between #M_(BC)# and S is
#d2=sqrt((2-3.5)^2+(8-3.5)^2)=sqrt(2.25+20.25)=sqrt(22.5)= 1.5*sqrt(10)#

Combining the equations [c] and [1]

#{y=(3x)/4+3.375#
#{x=2#
=> #y=1.5+3.375# => #y=4.875#

We've found #T(2, 4.875)#
The distance between #M_(CA)# and T is
#d3=sqrt((2-3.5)^2+(4.875-6)^2)=sqrt(2.25+1.265625)=sqrt(3.515625)=1.875#

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