Question

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)
If true then enter $1$ and if false then enter $0$

Answer 1

$Letustaketheco-ordinatesofthecircumcenterofthe\Delta PQRwithverticesP(x_1,y_1)=(2,7),Q(x_2,y_2)=(-5,8)\&R(x_3,y_3)=(-6,1)asC(x,y).Thenx=[({x_1}^2+{y_1}^2)\left(y_2-y_3\right)+({x_2}^2+{y_2}^2)\left(y_3-y_1\right)+({x_3}^2+{y_3}^2)\left(y_1-y_2\right)]\times \frac{1}{D}whenD=2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]andy=[({x_1}^2+{y_1}^2)\left(x_3-x_2\right)+({x_2}^2+{y_2}^2)\left(x_1-x_3\right)+({x_3}^2+{y_3}^2)\left(x_2-x_1\right)]\times \frac{1}{D}.\therefore D=2[2(8-1)+(-5)(1-7)+(-6)(7-8)]=2(14+30+6)=100andx=[(2^2+7^2)(8-1)+(5^2+8^2)(1-7)+(6^2+1^2)(7-8)+]\times \frac{1}{D}=(371-534-37)\times \frac{1}{100}=-2.Alsoy=[(2^2+7^2)(-6+5)+(5^2+8^2)(2+6)+(6^2+1^2)(-5-2)+]\times \frac{1}{D}=(-53+712-259)\times \frac{1}{100}=4.\therefore C(x,y)=(-2,4)Ans-Thecircumcentreof\Delta PQRisC(x,y)=(-2,4).$