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 betweenM_(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 betweenM_(BC) andS 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 betweenM_(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