Question

find circumcentre of a triangle whose sides are x=1,y=1,x+y=1​

Answer 1

Answer:

3 units

Step-by-step explanation:

x+y+x+y

=1+1+1

=3 units

Answer 2

Answer:

( 1/2 , 1/2)

Step-by-step explanation:

Let, (x,y) be the coordinates of the circumcenter

D1 be the distance from the circumcenter to vertex A

D2 be the distance from the circumcenter to vertex B

D3 be the distance from the circumcenter to vertex C

Since , the sides of the triangle is given as x=1 , y=1 , x+y=1​.

Now, we just want to find the coordinates of the points of the triangle.

Since x = 1 and y=1 are gven , ∴ the two sides will be A (0 , 1)  ,  B (1 , 0) and the third point will come from the line x+y=1

So, the points will be A (0 , 1)  ,  B (1 , 0)   ,  C (1 , 1)

$D1 = \sqrt{(X-0)^{2} +(Y-1)  ^{2}  }$

$D1 = \sqrt{(X)^{2} +(Y-1)  ^{2}  }$............(1)

$D2= \sqrt{(X-1)^{2} +(Y-0)  ^{2}  }$

$D2 = \sqrt{(X-1)^{2} +(Y)  ^{2}  }$.............(2)

$D3 = \sqrt{(X-1)^{2} +(Y-1)  ^{2}  }$..........(3)

Since, the distance of vertex and circumcenter are same,

So, D1 = D2

we will get

X = Y............(4)

Making D1 = D3,

we wil get,

$X^{2}  + (Y-1)^{2}  = X^{2}  +  Y^{2}  - 2  \times X - 2 \times Y +2$

By solving above, we get;

X= 1/2

Put X in eq (4)

We get;

Y = 1/2

So, the Circumcentre of a triangle is (1/2 , 1/2)