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

1 Answer
Jan 15, 2016

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.

Explanation:

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

So
line 1 perpendicular to AB meets side BC
line 2 perpendicular to BC meets side AB
line 3 perpendicular to AC meets side AB

We need the equations of the lines in which the sides AB and BC lay and the equations of the 3 perpendicular lines

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]

Finding the interceptions on sides AB and BC

Combining equations [b] and [1]

{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

Combining equations [a] and [2]

{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)

Combining the equations [a] and [3]

{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