Question

The sides of a triangle are $x=2,y+1=0$ and $x+2y=4$. Its circumcentre is-

• A. $(4,0)$
• B. $(2,-1)$
• C. $(0,4)$
• D. $(2,3)$

Answer 1

Answer: (A)

$(4,0)$

sides are $x=2,y=-1.x+2y=4$

Solving these we get get vertices of the

Triangle are $A(2,-1),B(2,1),C(6,-1)$

Let circumcentre is $d(x,y)$

circumcentre is equidistant from all three vertices so

$C^{\prime }A=C^{\prime }B$

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

$y=0$

$C^{\prime }B=C^{\prime }C$

$(x-2{)}^2+(y-1{)}^2=(x-6{)}^2+(y+1{)}^2$

$Puty=0$

$(x-2{)}^2+1=(x-6{)}^2+1$

$x-2=\pm (x-6)$

$x-2=x-6$

Invalid

$x-2=-(x-6)$

$2x=8$

$x=4$

So, circumcentre is $C^{\prime }(4,0)$