Question

Find the circumvented of the triangle whose sides are x=1,y=1 and x+y=1

Answer 1

( 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)

BTS AND EXO