Question

Find the coordinates of the circumcenter of △PQR with vertices P(−2,5), Q(4,1), and R(−2,−3).

Answer 1

Answer: (1/3,2)

Step-by-step explanation:

Let O(x,y) be the circumference of the triangle PQR,

Thus, by the property of circumcenter all the vertices of the triangle are at same distance from this circumcenter.

Thus, PO = QO

By the distance formula,

$(x+2)^2+(y-5)^2 = (x-4)^2+(y-1)^2$

$\impliesx^2+ 4x + 4 + y^2+ 25 - 10 y = x^2+16 - 8 x + y^2+1 - 2y$

$\implies 4x + 4 + 25 - 10 y = 16 - 8 x + 1 - 2y$

$\implies 4x +8x-10y+2y = 16 + 1 -25-4$

$\implies 12x - 8y = 17 - 29$

$\implies 12 x - 8 y = - 12$---------(1)

Similarly, QO = RO

$\implies (x-4)^2 + (y-1)^2 = (x+2)^2+(y+3)^2$

$\implies x^2 + 16 - 8x + y^2+ 1 - 2y = x^2+ 4 + 4 x +y^2+ 9 + 6 y$

$\implies 16 - 8x + 1 - 2y = 4 + 4 x + 9 + 6 y$

$\implies - 8x -4x-2y-6y=4+9-16-1$

$\implies -12 x - 8 y = 13 - 17$

$\implies 12 x + 8 y = 4$   ------------(2)

By adding equation (1) and (2),

24 x = - 8

x = 1/3

By putting this value in equation (1),

We get,

Thus, 4 - 8 y = -12

8 y = 16

y = 2