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Find the circumcentre of the triangle whose vertices are (-2, -3), (- 1, 0), (7, - 6). |
Answer» Vertices of triangle are A(-2, -3), B(- 1, 0), C(7, - 6) Let the coordinates of P are (x, y) PA = PB = PC PA = PB \(\sqrt{(x + 2)^2 + (y + 3)^2}\) = \(\sqrt{(x + 1)^2 + (y + 0)^2}\) On squaring both sides, we get x2 + 4x + 4 +y2 +6y + 9 = x2 + 2x + 1 + y2 2x + 6y = - 12 x + 3y = - 6.......(1) PA = PC \(\sqrt{(x + 2)^2 + (y + 3)^2}\) = \(\sqrt{(x - 7)^2 + (y + 6)^2}\) On squaring both sides, we get x2 + 4x + 4 + y2 + 6y + 9 = x2 - 14x + 49 + y2 + 12y + 36 18x - 6y = 72 3x - y = 12.......(2) On solving equations (1) & (2), We get x = 3 and y = -3 Therefore coordinates are (3, -3) |
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