Question

Determine the coordinates of the circumcentre of triangle whose vertices are (12,5),(6,-1)&(12,-1)

Answer 1

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Answer 2

Answer:

The coordinates of the circumcentre = (9,2)

Explanation:

Given,

The vertices of the triangle are  (12,5),  (6,-1), (12,-1)

Let ABC be the triangle and the circumcentre be 'O'.

The coordinates of the triangle are A(12,5), B(6,-1), and C(12,-1).

and let the coordinates of the circumcentre be O(x,y)

We know,

All the vertices of a triangle are equidistant from the circumcenter.

Then we have

OA = OB = OC -----------(1)

We know,

The distance between two points P$(x_1,y_1)$ and Q$(x_2,y_2)$ is given by the formula

PQ = $\sqrt{(x_1 - x_2)^2-(y_1 - y_2)^2}$

Substituting A(12,5), B(6,-1) C(12,-1) and O(x,y) we get

OA = $\sqrt{(x-12)^2 + (y - 5)^2}$

OB = $\sqrt{(x -6)^2 +(y+1)^2}$

OC = $\sqrt{(x- 12)^2 + (y +1)^2}$

From equation(1) we get

OA = OB

$\sqrt{(x-12)^2 + (y - 5)^2}$   = $\sqrt{(x -6)^2 +(y+1)^2}$

$(x-12)^2 + (y - 5)^2$ = $(x -6)^2 +(y+1)^2$

Expand the terms using the identity$(a+b)^2 = a^2 +2ab +b^2$ we get

$x^2 - 24x +144 +y^2 - 10y +25 = x^2 -12x +36 +y^2 +2y +1$

-24x +12x -10y -2y = 36+1 -25-144

-12x -12y = -132

x+y = 11 ----------------(2)

similarly from equaton (1) we get,

OA = OC

$\sqrt{(x-12)^2 + (y - 5)^2}$   = $\sqrt{(x - 12)^2 + (y+1)^2}$

$(x-12)^2 + (y - 5)^2$ = $(x - 12)^2 + (y+1)^2$

Expand the terms using the identity$(a+b)^2 = a^2 +2ab +b^2$ we get

$x^2 -24x +144 + y^2 -10y +25 = x^2 -24x +144 +y^2 +2y +1$

-24x +24x -10y -2y = 144+1 -144 -25

-12y = -24

y = 2 ----------(3)

Substituting the value of 'y' in equation (2) we get,

x+2 = 11

x =9

Hence,

the coordinates of the circumcentre = (x,y) = (9,2)