find the circumcentre of the triangle whose vertices are (-2,-3), (-1,0), (7,-6) - Maths - Coordinate Geometry

Question

find the circumcentre of the triangle whose vertices are
(-2,-3), (-1,0), (7,-6)

Chetna Khanna · Accepted Answer

Circumcentre of a triangle is the point of intersection of all the three perpendicular bisectors of the sides of triangle.

So, the vertices of the triangle lie on the circumference of the circle.

Let the coordinates of the circumcentre of the triangle be (x, y).

(x, y) will the equidistant from the vertices of the triangle.

Using distance formula, $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ , it is obtained:

$D_{1} = \sqrt{(x + 2)^{2} + (y + 3)^{2}} \Rightarrow D_{1} = \sqrt{x^{2} + 4 + 4 x + y^{2} + 9 + 6 y} (Taking points (x, y) and (–2, –3)) \Rightarrow D_{1} = \sqrt{x^{2} + y^{2} + 4 x + 6 y + 13} D_{2} = \sqrt{(x + 1)^{2} + (y - 0)^{2}} (Taking points (x, y) and (–1, 0)) \Rightarrow D_{2} = \sqrt{x^{2} + 1 + 2 x + y^{2}} D_{3} = \sqrt{(x - 7)^{2} + (y + 6)^{2}} (Taking points (x, y) and (7, –6)) \Rightarrow D_{3} = \sqrt{x^{2} + 49 - 14 x + y^{2} + 36 + 12 y} \Rightarrow D_{3} = \sqrt{x^{2} + y^{2} - 14 x + 12 y + 85}$

As (x, y) is equidistant from all the three vertices.

So, D₁ = D₂ = D₃

D₁ = D₂

$\Rightarrow \sqrt{x^{2} + y^{2} + 4 x + 6 y + 13} = \sqrt{x^{2} + 1 + 2 x + y^{2}} \Rightarrow x^{2} + y^{2} + 4 x + 6 y + 13 = x^{2} + 1 + 2 x + y^{2} \Rightarrow 4 x + 6 y - 2 x = 1 - 13 \Rightarrow 2 x + 6 y = - 12 \Rightarrow x + 3 y = - 6 ... (1) D_{2} = D_{3} \Rightarrow \sqrt{x^{2} + 1 + 2 x + y^{2}} = \sqrt{x^{2} + y^{2} - 14 x + 12 y + 85} \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + y^{2} - 14 x + 12 y + 85 \Rightarrow 2 x + 14 x - 12 y = 85 - 1 \Rightarrow 16 x - 12 y = 84 \Rightarrow 4 x - 3 y = 21 ... (2)$

Adding equations (1) and (2):

$x + 3 y + 4 x - 3 y = - 6 + 21 \Rightarrow 5 x = 15 \Rightarrow x = \frac{15}{5} \Rightarrow x = 3 and 4 x - 3 y = 21 \Rightarrow 3 y = 4 x - 21 \Rightarrow y = \frac{4 x - 21}{3} \Rightarrow y = \frac{4 (3) - 21}{3} \Rightarrow y = \frac{12 - 21}{3} \Rightarrow y = - \frac{9}{3} \Rightarrow y = - 3$

∴ (3, –3) are the coordinates of the circumcentre of the triangle.