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