Question

Find the circumcentre of the triangle whose vertices are (1,0) (-1,2) (3,2)

Answer 1

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).

Answer 2

(1 , 2) is the circumcenter of the triangle whose vertices are (1,0) (-1,2) (3,2)

Given : A triangle whose vertices are (1,0) (-1,2) (3,2)

To Find : Circumcentre of the triangle

Solution:

"Circumcentre of the triangle is at equal distance from all the vertices"

Let say circumcenter is (x , y)

Distance from Each vertex are equal hence Square of Distance will also be equal.

=> (x - 1)² + (y - 0)² = (x - (-1))² + ( y - 2)² = (x - 3)² + ( y - 2)²

=> (x - 1)² + y² = (x + 1)² + ( y - 2)² = (x - 3)² + ( y - 2)²

Equating

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

=> (x + 1)² =  (x - 3)²

=> x² + 2x + 1 = x² -6x + 9

=> 8x = 8

=> x = 1

Equating

(x - 1)² + y² = (x + 1)² + ( y - 2)²

=> x² - 2x + 1 + y² = x² + 2x + 1  + y² - 4y  + 4

=>    0 = 4x - 4y + 4

x = 1

=> 0 = 4 - 4y + 4

=> 4y = 8

=> y = 2

Circumcentre of the triangle whose vertices are (1,0) (-1,2) (3,2) is ( 1 ,2 )

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