Question

Find the circumcentre of a triangle whose sides are x = 2, y = 2 and x + y = 0?

Answer 1

Answer: circumcenter is 0,0

Step-by-step explanation:

Points of intersection of the three sides are

A(2,2) ; B(2,-2) ; C(-2,2)

These are the vertices of the triangle formed by the three lines

The circumcenter of a triangle is the point which is equidistant from all three vertices of the triangle

Let the circumcenter coordinates be (x,y)

Distance formula = $\sqrt{ (x2-x1)^{2} + (y2-y1)^{2} }$

Where x1,y1 and x2,y2 are the vertices

Using this formula and substituting the values, we get

$\sqrt{ (x-2)^{2} + (y-2)^{2} } = \sqrt{ (x-2)^{2} + (y+2)^{2} } = \sqrt{ (x+2)^{2} + (y-2)^{2} }$

Upon squaring, we get

$=> (x-2)^{2} + (y-2)^{2} = (x-2)^{2} + (y+2)^{2} = (x+2)^{2} + (y-2)^{2}$

If we solve any two of them, we get our x and y values

$=> (x-2)^{2} + (y-2)^{2} = (x-2)^{2} + (y+2)^{2}\\\\=> (y-2)^{2} = (y+2)^{2}\\\\=> y^{2} +4 -4y = y^{2} + 4 + 4y\\\\=> 8y = 0\\\\=> y = 0$

Similarly x = 0

Therefore the circumcenter is 0,0

Please brainlist my answer, if helpful!