Question

The circumcentre of the triangle formed by the lines 3x-y-5=0,x+3y-5=0, x=y is

• A. (3, 1)
• B. (5/2, 5/2)
• C. (5/4, 5/4)
• D. (15/8, 15/8)

Answer 1

The correct option is A (3, 1)

$Formulaused$
$midpointformula=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$equationoflinepassingthroughtwopoints$
$y-y_1=m(x-x_1)$
$Given$
$3x-y=5.........\left(i\right)$
$x+3y=5..........\left(ii\right)$
$x=y...........\left(iii\right)$
$from\left(i\right)and\left(ii\right)$
$A(2,1)$
$from\left(ii\right)and\left(iii\right)$
$B(1.25,1.25)$
$from\left(i\right)and\left(iii\right)$
$C(2.5,2.5)$
$LetD,E,Fbethemidpo\mathrm{int}sofAB,BCandCArespectively$
$D(1.625,1.125)$
$E(1.875,1.875)$
$F(2.25,1.75)$
$Slopeof$
$\left(i\right)=3$
$\left(ii\right)=1/3$
$\left(iii\right)=1$
$Slopeoflineperpendiculatto$
$\left(i\right)=-1/3$
$\left(ii\right)=3$
$\left(iii\right)=-1$
$Equationofline$
$DO$
$(y-1.625)=\frac{-1}{3}(x-1.125)$
$\Rightarrow 3y-4.875=1.125-x$
$\Rightarrow 3y+x-6=0.......\left(iv\right)$
$SimilarlyequationofFO$
$y+x-4=0.........\left(v\right)$
$Solving\left(iv\right)and\left(v\right)$
$O(3,1)$