The circumcentre of the triangle formed by the lines $x+$ $y+2=0,2 x+y+8=0$ and $x-y-2=0$ is
The circumcentre of the triangle formed by the lines $x+$ $y+2=0,2 x+y+8=0$ and $x-y-2=0$ is
$(-5,1)$
$(-4,0)$
$(0,-2)$
$\left(\frac{-8}{3}, \frac{-2}{3}\right)$
Solution
$x+y+2=0$ $\qquad ....\mathrm{(i)}$
$\begin{aligned} & 2 x+y+8=0 \qquad ....\mathrm{(ii)}\\ & x-y-2=0 \qquad ....\mathrm{(iii)} \end{aligned}$
Solving (i) and (ii), we get
$(x, y)=(-6,4)$
Solving (i) and (iii). we get
$(x, y)=(0,-2)$
Solving (ii) and (iii), we get
$(x, y)=(-2,-4)$
Hence, vertices of triangles are $(0,-2),(-6,4)$,
$(-2,-4)$
Which is a right angled triangle,
Ends of hypotenuse are $(-6,4)$ and $(-2,-4)$ So, circumcentre = mid-point of hypotenuse $=\left(\frac{-6-2}{2}, \frac{4-4}{2}\right)=(-4,0)$
