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
Endpoints at pairs of coordinates [
Explanation:
Repeating the points
Midpoints
Slopes of segments (
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.
=>
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:
Equation of the line (passing through midpoint) perpendicular to side:
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#