The circumcenter of a triangle ABC whose vertices are A(0,0) B(6,0)and C(0,8) is
Answers
Step-by-step explanation:
Given:-
a triangle ABC whose vertices are A(0,0) B(6,0)and C(0,8)
To find:-
The circumcenter of a triangle ABC whose vertices are A(0,0) B(6,0)and C(0,8)
Solution:-
Circumcentre of a triangle is the point of intersection of all the three perpendicular bisectors of the sides of triangle.
So, the vertices of the triangle lie on the circumference of the circle.
Let the coordinates of the circumcentre of the triangle be P(x, y).
P(x, y) will the equidistant from the vertices of the triangle.
We know that
The distance between the two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units
I)Distance between P and A:-
(x1, y1)=(x,y)=>x1=x and y1=y
(x2, y2)=(0,0)=>x2=0 and y2=0
PA =√[(0-x)^2+(0-y)^2]
=>PA =√(x^2+y^2) units ------------(1)
ii)Distance between P and B:-
(x1, y1)=(x,y)=>x1=x and y1=y
(x2, y2)=(6,0)=>x2=6and y2=0
PB =√[(6-x)^2+(0-y)^2]
=>PB =√(36+x^2-12x+y^2) units --------(2)
iii) Distance between P and C:-
(x1, y1)=(x,y)=>x1=x and y1=y
(x2, y2)=(0,8)=>x2=0 and y2=8
PC =√[(0-x)^2+(8-y)^2]
=>PC=√(x^2+64-16y+y^2) units --------(3)
We have
PA=PB=PC
Now ,on taking PA = PB
=>√(x^2+y^2) = √(36+x^2-12x+y^2)
On squaring both sides then
=>[√(x^2+y^2)]^2 = [√(36+x^2-12x+y^2)]^2
=>x^2+y^2 = 36+x^2-12x+y^2
=>x^2+y^2-36-x^2+12x-y^2 = 0
=>-36+12x = 0
=>12x = 36
=>x = 36/12
=>x = 3 ----------(4)
On taking PB=PC
√(x^2+64-16y+y^2) = √(x^2+64-16y+y^2)
On squaring both sides then
=>[(36+x^2-12x+y^2)]^2= [√(x^2+64-16y+y^2)]^2
=> 36+x^2-12x+y^2= x^2+64-16y+y^2
=>36+x^2-12x+y^2-x^2-64+16y-y^2 = 0
=>36-12x-64+16y = 0
=>-12x+16y-28 = 0
=>-12(3)+16y-28=0
=>-36+16y-28 = 0
=>-64+16y = 0
=>16y = 64
=>y = 64/16
=>y = 4
P(x,y)=(3,4)
Answer:-
The circumcenter of the given triangle is (3,4)
Used formulae:-
- Circumcentre of a triangle is the point of intersection of all the three perpendicular bisectors of the sides of triangle.
- The vertices of the triangle lie on the circumference of the circle.
- The distance between the two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units
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hope it helps....
