Question

The vertices of a triangle are (5,1) (11,1) and (11,9). Find the co-ordinates of the circumcenter of the triangle.

Answer 1

Vertices of the triangle are A(5,1), B(11,1) and C (11,9).

Let P(x, y) be the circumcenter of the triangle.

So, PA = PB = PC (Circumcenter is equidistant from the vertices of the triangle)
=>(PA)^2 = (PB)^2

=>{(x – 5)^2 + (y – 1)^2} = {(x –11)^2 + (y - 1)^2}
=>x^2 + 25 – 10x + y^2 + 1 – 2y = x^2 + 121 – 22x + y^2 + 1 - 2y
=> 12x = 96
=> x = 8.........(1)


As,(PA)^2 = (PC)^2
=>{ (x –5)^2 + (y – 1)^2 } = { (x – 11)^2 + (y - 9)^2 }
=> x^2 + 25 – 10x + y^2 +1 –2y = x^2 + 121 – 22x + y^2 + 81 - 18y
=> 25+1 -10x-2y = 121+81-22x -18y
=> 26-10x-2y= 202 -22x-18y
=> 22x-10x +18y-2y = 202 -26
=> 12x +16y = 276
=> 12(8) +16y = 276 .....(using 1)
=> 16y = 276-96
=> 16y = 180
=> y = 180/16
=>y = 45/4 so coordinates of circumcenter is (8,45/4)
hope it helps !