The circumcentre of the triangle with vertices at ( - 2 ,   3 ) ,   ( 1 , - 2 )   and ( 2 , 1 ) is

Question

The circumcentre of the triangle with vertices at ( - 2 , &#160; 3 ) , &#160; ( 1 , - 2 ) &#160; and ( 2 , 1 ) is

Aicharya · Accepted Answer

Circumcentre of the triangle is equidistant from the vertices of the triangle. Let $O (x, y)$ be the circumcentre of the triangle.

Now,

$O A = \sqrt{{(x + 2)}^{2} + {(y - 3)}^{2}} . . . (i)$

$O B = \sqrt{{(x - 1)}^{2} + {(y + 2)}^{2}} . . . (ii)$

$O C = \sqrt{{(x - 2)}^{2} + {(y - 1)}^{2}} . . . (iii)$

$O A = O B = O C$

Equating $(i) & (ii)$ , we get

$\sqrt{{(x + 2)}^{2} + {(y - 3)}^{2}} = \sqrt{{(x - 1)}^{2} + {(y + 2)}^{2}}$

$\Rightarrow {(x + 2)}^{2} + {(y - 3)}^{2} = {(x - 1)}^{2} + {(y + 2)}^{2}$

$3 x - 5 y + 4 = 0 . . . (iv)$

Equating $(ii) & (iii)$ , we get

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

$\Rightarrow {(x - 1)}^{2} + {(y + 2)}^{2} = {(x - 2)}^{2} + {(y - 1)}^{2}$

$x + 3 y = 0 . . . (v)$

On solving $(iv) & (v)$ , we get

$(x, y) \equiv (\frac{- 6}{7}, \frac{2}{7})$

The circumcentre of the triangle with vertices at ( - 2 ,   3 ) ,   ( 1 , - 2 )   and ( 2 , 1…