find the circum center of the triangle whose vretices are given below (1,3)(0,-2)(-3,1) solution case 1 case2
Answers
Answer:
(1/3,2/3)
Step-by-step explanation:
Using distance formula, we can find circum center of a triangle
It is nothing but is the point of intersection of all the three perpendicular bisector of sides of A
Let co ordinates of circum center be (x,y) distance formula is used
D
1
using point (−2,3)
D
1
=
(x+2)
2
+(y−3)
2
=
x
2
+4x+4+y
2
−6y+9
=
x
2
+y
2
+4x−6y+13
=D
2
using point (2,1)
D
2
=
(x−2)
2
+(y+1)
2
=
x
2
+4x+4+y
2
+2y+1
=
x
2
+y
2
+4x+2y+5
D
3
using point (4,0)
D
3
=
(x−4)
2
+(y−0)
2
=
x
2
−8x+16+y
2
=
x
2
+y
2
−8x+16
As (x,y) is equidistant from all three vertices
D
1
=D
2
=D
3
∴D
1
=D
2
∴
x
2
+y
2
+4x−6y+13
=
x
2
+y
2
+4x+2y+5
∴x
2
+y
2
+4x−6y+13=x
2
+y
2
+4x+2y+5
∴−6y+13=2y+5
∴8=8y
∴y=1
D
1
=D
3
∴
x
2
+y
2
+4x−6y+13
=
x
2
+y
2
−8x+16
∴x
2
+y
2
+4x−6y+13=x
2
+y
2
−8x+16
∴12x=3+6y
But y=1
∴x=
12
9
=
4
3
∴Circum center is (
4
3
,1)