A triangle has corners at #(6 , 1 )#, ( 4, 2 )#, and #( 2, 8 )#. What are the endpoints and lengths of the triangle's perpendicular bisectors?
1 Answer
Endpoints at [
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 [c]
line [2] perpendicular to BC meets side CA [c]
line [3] perpendicular to AC meets side BC [b]
We need the equations of the lines in which the sides BC 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 BC and CA
Combining equations [1] and [c]
#{y=2x-8.5#
#{y=(-7x+46)/4# =>#2x-8.5=(-7x+46)/4# =>#8x+34=-7x+46# =>#15x=80# =>#x=16/3#
#-> y=2*16/3-17/2=(96-51)/6=45/6# =>#y=15/2# We've found
#R(16/3,7.5)#
The distance between#M_(AB)# and R is
#d1=sqrt((16/3-5)^2+(15/2-3/2)^2)=sqrt((1/3)^2+6^2)=sqrt(1+324)/3=sqrt(325)/3=(5*sqrt(13))/3~=6.009#
Combining equations [2] and [c]
#{y=(x+12)/3#
#{y=(-7x+46)/4# =>#(x+12)/3=(-7x+46)/4# =>#4x+48=-21x+138# =>#25x=90# =>#x=18/5#
#-> y=(18/5+12)/3=(18+60)/15# =>#y=26/5# We've found
# S(3.6,5.2)#
The distance between#M_(BC)# and#S# is:
#d2=sqrt((18/5-3)^2+(26/5-5)^2)=sqrt(3^2+1^2)/5=sqrt(10)/5~=.632#
Combining the equations [3] and [a]
#{y=(4x+15.5)/7#
#{y=-3x+14# =>#(4x+15.5)/7=-3x+14# =>#4x+15.5=-21+98# =>#25=82,5# =>#x=3.3#
#-> y=-3*3.3+14# =>#y=4.1# We've found
#T(3.3,4.1)#
The distance between#M_(CA)# and T is
#d3=sqrt((3.3-4)^2+(4.1-4.5)^2)=sqrt(.7^2+.4^2)=sqrt(.49+.16)=sqrt(.65)=sqrt(65)/10~=.806#