Question

5.Find the circumcentre of the triangle whose vertices are (1 3) (-3 5) and (5 1)
pls answer and don't waste my time​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

❖ Let us consider a triangle ABC having vertices A(1, 3), B(-3, 5) and C(5, 1) respectively.

❖ Let O(x, y) be the circumcentre of triangle ABC.

⇛ OA = OB = OC

$\tt \:  ⟼ Consider \: OA \: = \: OB$

$\bf\implies \: {OA }^{2} = {OB}^{2}$

$\tt \:  ⟼ {(x - 1)}^{2} + {(y - 3)}^{2} = {(x + 3)}^{2} + {(y - 5)}^{2}$

$\tt \:  \cancel{x}^{2} + 1 - 2x + \cancel{y}^{2} + \cancel9 - 6y = \cancel{x}^{2} + \cancel9 + 6x + \cancel{y}^{2} + 25 + 10y$

$\tt \:  ⟼ 8x + 16y = - 24$

$\bf\implies \:x + 2y = - 3$

$\tt\implies \:x = - 3 - 2y - - - - (1)$

$\tt \:  ⟼ Consider \: OB \: = \: OC$

$\bf\implies \: {OB}^{2} = {OC}^{2}$

$\tt \:  ⟼ {(x + 3)}^{2} + {(y - 5)}^{2} = {(x - 5)}^{2} + {(y - 1)}^{2}$

$\tt \:  \cancel{x}^{2} + 9 + 6x + \cancel{y}^{2} + \cancel25 - 10y = \cancel{x}^{2} + \cancel25 - 10x + \cancel{y}^{2} + 1 - 2y$

$\tt\implies \:16x - 8y = - 8$

$\tt\implies \:2x - y = - 1$

$\tt\implies \:2( - 3 - 2y) - y = - 1 \: \: (using \: (1))$

$\tt\implies \: - 6 - 4y - y = - 1$

$\tt\implies \:5y = - 5$

$\tt\implies \:y = - 1$

☆ On substituting the value of y in equation (1), we get

$\tt \:  ⟼ x = - 3 - 2( - 1)$

$\tt\implies \: x = - 3 + 2$

$\tt\implies \:x = - 1$

$\bf\implies \:So, \: coordinates \: of \: circumcentre \: is \: (-1,-1)$