Question #139730
5)The coordinates of the vertices of a triangle are (3,2), (9,2), (6,5).
a)Find the equations of the perpendicular bisectors of triangle.
b)Find the co ordinates of the circumcentre
1
Expert's answer
2020-10-26T19:01:32-0400

The coordinates of the vertices of a triangle are A(3,2),B(9,2),C(6,5).A(3,2), B(9,2), C(6,5).

a)

The midpoint of ABAB is M1(3+92, 2+22)=M1(6, 2)M_1\big(\frac{3+9}{2},\ \frac{2+2}{2}\big)=M_1(6,\ 2)

The slope of ABAB is 2293=06=0\frac{2-2}{9-3}=\frac{0}{6}=0. It means that ABAB is a horizontal line and perpendicular bisector is a vertical line. The equation of the perpendicular bisector is x=6x=6 .


The midpoint of BCBC is M2(9+62, 2+52)=M2(7.5, 3.5)M_2\big(\frac{9+6}{2},\ \frac{2+5}{2}\big)=M_2(7.5,\ 3.5) 

The slope of BCBC is 2596=33=1\frac{2-5}{9-6}=\frac{-3}{3}=-1. The slope of the perpendicular bisector is m2=11=1m_2=-\frac{1}{-1}=1 .

The equation of the perpendicular bisector is y3.5=1×(x7.5)y-3.5=1\times (x-7.5) or y=x4y=x-4 .


The midpoint of ACAC is M3(3+62, 2+52)=M3(4,5, 3.5)M_3\big(\frac{3+6}{2},\ \frac{2+5}{2}\big)=M_3(4,5,\ 3.5) 

The slope of ACAC is 2536=33=1\frac{2-5}{3-6}=\frac{-3}{-3}=1. The slope of the perpendicular bisector is m3=11=1m_3=-\frac{1}{1}=-1 .The equation of the perpendicular bisector is y3.5=1×(x4.5)y-3.5=-1\times (x-4.5) or y=x+8y=-x+8 .

b)

The circumcenter is the point of concurrency of perpendicular bisectors of a triangle. We need to solve system of any two bisector equations to find the coordinates of the circumcenter:

{x=6y=x4\begin{cases} x=6 \\ y=x-4 \end{cases} {x=6y=2\begin{cases} x=6 \\ y=2 \end{cases}

The circumcenter is O(6, 2)O(6,\ 2) .


Answers: a) x=6x=6 , y=x4y=x-4 , y=x+8y=-x+8 ; b) (6, 2)(6,\ 2) .



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