find the circumtance of a triangle ABC with virtices A(5,1) , B(-3,-7) and C(7,-11)
Answers
Answer:
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Step-by-step explanation:
The circumcentre of∆ABC is (2,-4)
Step-by-step explanation:
The vertices of ∆ABC are A(5,1) B(-3,-7) and C(7,-1)
Circumcentre: It is point of intersection of perpendicular bisector of each sides of triangle.
# Midpoint of side AB at P, =(\frac{5-3}{2},\frac{1-7}{2})=(
2
5−3
,
2
1−7
)
Midpoint of side AB at P, =(1,-3)=(1,−3)
Slope of line AB, m =\dfrac{-7-1}{-3-5}=1=
−3−5
−7−1
=1
Equation of perpendicular line bisector of AB passing through point P
y+3=\dfrac{-1}{1}(x-1)y+3=
1
−1
(x−1)
y=-x-2y=−x−2 ---------- (1)
# Midpoint of side BC at Q, =(\frac{7-3}{2},\frac{-1-7}{2})=(
2
7−3
,
2
−1−7
)
Midpoint of side BC at Q, =(2,-4)=(2,−4)
Slope of line BC, m =\dfrac{-7+1}{7+3}=-\dfrac{3}{5}=
7+3
−7+1
=−
5
3
Equation of perpendicular line bisector of BC passing through point Q
y+4=-\dfrac{5}{3}(x-2)y+4=−
3
5
(x−2)
y=-\dfrac{5}{3}x-\dfrac{2}{3}y=−
3
5
x−
3
2
------------(2)
Point of intersection of equation (1) and equation (2) is circumcentre of triangle.
Using substitution method:-
-\dfrac{5}{3}x-\dfrac{2}{3}=-x-2−
3
5
x−
3
2
=−x−2
x=2x=2
Put x=2 into y=-x-2
y=-4
Cirncumcentre: (2,-4)