1.

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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