Question

The circumcentre of the triangle formed by the lines $2x^2-3xy-2y^2=0$ and $3x-y=10$ is

• A. $(2,1)$
• B. $(1,-2)$
• C. $(3,-6)$
• D. $(3,-1)$

Answer 1

Answer: (D)

$(3,-1)$

$2x^2-3xy-2y^2=0$

Substitute $t=\frac{x}{y},$

$2t^2-3t-2=0$

$\Rightarrow t=\frac{3\pm \sqrt{9+16}}{4}$

$\Rightarrow t=2,\frac{-1}{2}$

$\Rightarrow y=\frac{x}{2}$ & $y=-2x$

And $3x-y=10$

For circumcentre A,B,C are lie on a circle.

So, $\angle BAC={90}^{\circ }$

BC is a diameter of that circle

So, centre of circle is circumcentre

$\therefore$ centre is mid point of BC

Circumcentre is $(\frac{6}{2},\frac{-2}{1})=(3,-1).$