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Finding the Coordinates of a Point

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Question

Find the co-ordinates of the circumcentre of triangle if P $\equiv$ (2,7), Q $\equiv$ (-5, 8), R $\equiv$ (-6, 1) are the vertices of triangle.

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Solution

$Let the coordinate s of the circumcentre O be (x, y) . Distance of circumcentre from the vertices P (2, 7), Q (- 5, 8) and R (- 6, 1) of the triangle will be equal, as it is equal to the radius of the circle .$

$∴ OP = OQ \Rightarrow {OP}^{2} = {OQ}^{2} \Rightarrow {(x - 2)}^{2} + {(y - 7)}^{2} = {[x - (- 5)]}^{2} + {(y - 8)}^{2} \Rightarrow {(x - 2)}^{2} + {(y - 7)}^{2} = {(x + 5)}^{2} + {(y - 8)}^{2} \Rightarrow x^{2} + 4 - 4 x + y^{2} + 49 - 14 y = x^{2} + 25 + 10 x + y^{2} + 64 - 16 y \Rightarrow - 4 x - 14 y + 53 = 10 x - 16 y + 89 \Rightarrow - 14 x + 2 y = 36 \Rightarrow - 7 x + y = 18 - - - - (1)$

$∴ OP = OR \Rightarrow {OP}^{2} = {OR}^{2} \Rightarrow {(x - 2)}^{2} + {(y - 7)}^{2} = {[x - (- 6)]}^{2} + {(y - 1)}^{2} \Rightarrow {(x - 2)}^{2} + {(y - 7)}^{2} = {(x + 6)}^{2} + {(y - 1)}^{2} \Rightarrow x^{2} + 4 - 4 x + y^{2} + 49 - 14 y = x^{2} + 36 + 12 x + y^{2} + 1 - 2 y \Rightarrow - 4 x - 14 y + 53 = 12 x - 2 y + 37 \Rightarrow - 16 x - 12 y = - 16 \Rightarrow 4 x + 3 y = 4 - - - - (2)$

$Multiplying equation (1) by 3, we get : - 21 x + 3 y = 54 - - - - (3) Subtracting equation (2) from (3), we get : - 21 x + 3 y = 54 4 x + 3 y = 4 - - - - 25 x = 50 \Rightarrow x = - 2 Substituting the value of x in equation (1), we get : - 7 \times (- 2) + y = 18 \Rightarrow 14 + y = 18 \Rightarrow y = 4 Therefore, (- 2, 4) are the coordinate s of the circumcentre of triangle PQR .$

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Q. The co-ordinates of the circumcentre of triangle if

P \equiv (2, 7), Q \equiv (- 5, 8), R \equiv (- 6, 1)

are the vertices of triangle is (-2, 4)
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R (x_{3}, y_{3})

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