Find the circumcentre of the triangle whose

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be A(−2.−3), B(−1,0) and C(7,−6).

Let the circumcentre of the triangle be represented by the point R(x, y).

So we have $A R=B R=C R$

$A R=\sqrt{(-2-x)^{2}+(-3-y)^{2}}$

$B R=\sqrt{(-1-x)^{2}+(-y)^{2}}$

$C R=\sqrt{(7-x)^{2}+(-6-y)^{2}}$

Equating the first pair of these equations we have,

$A R=B R$

$\sqrt{(-2-x)^{2}+(-3-y)^{2}}=\sqrt{(-1-x)^{2}+(-y)^{2}}$

Squaring on both sides of the equation we have,

$(-2-x)^{2}+(-3-y)^{2}=(-1-x)^{2}+(-y)^{2}$

$4+x^{2}+4 x+9+y^{2}+6 y=1+x^{2}+2 x+y^{2}$

$2 x+6 y=-12$

$x+3 y=-6$

Equating another pair of the equations we have,

$A R=C R$

$\sqrt{(-2-x)^{2}+(-3-y)^{2}}=\sqrt{(7-x)^{2}+(-6-y)^{2}}$

Squaring on both sides of the equation we have,

$(-2-x)^{2}+(-3-y)^{2}=(7-x)^{2}+(-6-y)^{2}$

$4+x^{2}+4 x+9+y^{2}+6 y=49+x^{2}-14 x+36+y^{2}+12 y$

$18 x-6 y=72$

$3 x-y=12$

Now we have two equations for ‘x’ and ‘y’, which are

$x+3 y=-6$

$3 x-y=12$

From the second equation we have. Substituting this value of ‘y’ in the first equation we have,

$x+3(3 x-12)=-6$

$x+9 x-36=-6$

$10 x=30$

$x=3$

Therefore the value of ‘y’ is,

$y=3 x-12$

$y=3 x-12$

$y=-3$

Hence the co-ordinates of the circumcentre of the triangle with the given vertices are $(3,-3)$.