Question

Find the circumcentre of the triangle whose verteses are (1, 0), (–1, 2) and (3, 2).​

Answer 1

Answer:

10,-12,32

Step-by-step explanation:

where is your triangle

Answer 2

Answer:

(x,y)=(1,2)

Step-by-step explanation:

Given,

A(1,0) B(-1,2) C(3,2)

Let S(x,y) be the circumcenter

SA=SB=SC

Case--(1)

SA=SB

Squaring on both sides

SA^2=SB^2

(x-1)^2+(y-0)^2=(x+1)^2+(y-2)^2

x^2+1-2x+y^2=x^2+1+2x+y^2+4-4y

2x+4-4y+2x=0

4x-4y+4=0

4(x-y+1)=0

x-y+1=0

Case----(2)

SB=SC

Squaring on both sides

SB^2=SC^2

(x+1)^2+(y-2)^2=(x-3)^2+(y-2)^2

x^2+1+2y+y^2+4-4y=x^2+9-6x+4-4y

2x+4-4y=9-6x+4-4y

2x+1+6x-9=0

8x-8=0

8(x-1)=0

x-1=0

Point Of Intersection

-1 1 1 -1

0 -1 1 0

x/1-0 = y/1+1 =1/0+1

x/1=y/2=1/1

x=1 y=2

Therefore ,the circumcenter of S(x,y) is S(1,2).