Question

If the vertices of a triangle be (2, 1), (5, 2) and (3, 4) then its circumcenter is

• A. $(\frac{13}{2},\frac{9}{2})$
• B. $(\frac{13}{4},\frac{9}{4})$
• C. $(\frac{9}{4},\frac{13}{4})$
• D. None

Answer 1

The correct option is B $(\frac{13}{4},\frac{9}{4})$

$Given-theco-ordinatesofthecircumcenterofthe\Delta PQRwithverticesP(x_1,y_1)=(2,1),Q(x_2,y_2)=(5,2)\&R(x_3,y_3)=(3,4)areC(x,y).Tofindout-C(x,y)=?Solution-HerePA=PB=PCsincetheyaretheradiiofthecircumcentre.\therefore {PA}^2={PB}^2={PC}^2.Usingdistanceformula,{PA}^2={(x-2)}^2+{(y-1)}^2,{PB}^2={(x-5)}^2+{(y-2)}^2,{PC}^2={(x-3)}^2+{(y-4)}^2.{PA}^2={PB}^2.\therefore {(x-2)}^2+{(y-1)}^2={(x-5)}^2+{(y-2)}^2⟹3x+y-12=0⟹y=12-3x...........\left(i\right).Again{PB}^2={PC}^2.\therefore {(x-5)}^2+{(y-2)}^2={(x-3)}^2+{(y-4)}^2⟹-x+y+1=0⟹y=x-1........\left(ii\right).Compareing\left(i\right)\&\left(ii\right),12-3x=x-1⟹x=\frac{13}{4}.Puttingx=\frac{13}{4}in\left(ii\right),y=\frac{13}{4}-1=\frac{9}{4}.\therefore ThecircumcentreC(x,y)=(\frac{13}{4},\frac{9}{4}).Ans-OptionB.$