Question

Fund the Cirucmcenter of the triangle whose sides are x+y+2=0, 5x-y-2=0, x-2y+5=0

Answer 1

Let us find the three vertices of the triangle.

Find value of $y$ in $x+y+2=0$ and substitute it in $5x-y-2=0$ and $x-2y+5=0$ each.

So,

$\longrightarrow x+y+2=0$

$\longrightarrow y=-x-2$

And in,

$\longrightarrow 5x-y-2=0$

substituting value of $y,$

$\longrightarrow 5x-(-x-2)-2=0$

$\longrightarrow 5x+x+2-2=0$

$\longrightarrow6x=0$

$\longrightarrow x=0$

And then,

$\longrightarrow y=-x-2$

$\longrightarrow y=-0-2$

$\longrightarrow y=-2$

Thus (0, -2) is a vertex.

And in,

$\longrightarrow x-2y+5=0$

$\longrightarrow x-2(-x-2)+5=0$

$\longrightarrow x+2x+4+5=0$

$\longrightarrow 3x+9=0$

$\longrightarrow x=-3$

And then,

$\longrightarrow y=-x-2$

$\longrightarrow y=-(-3)-2$

$\longrightarrow y=1$

Thus (-3, 1) is a vertex.

Consider,

$\longrightarrow 5x-y-2=0$

$\longrightarrow y=5x-2$

And in,

$\longrightarrow x-2y+5=0$

substituting value of $y,$

$\longrightarrow x-2(5x-2)+5=0$

$\longrightarrow x-10x+4+5=0$

$\longrightarrow -9x+9=0$

$\longrightarrow x=1$

And then,

$\longrightarrow y=5x-2$

$\longrightarrow y=5(1)-2$

$\longrightarrow y=3$

Thus (1, 3) is a vertex of the triangle.

So (0, -2), (-3, 1) and (1, 3) are vertices of the triangle.

Let (a, b) be the coordinates of the circumcenter of the triangle, which is equidistant from the points (0, -2), (-3, 1) and (1, 3) each.

Then by distance formula,

$\longrightarrow\sqrt{(a-0)^2+(b+2)^2}=\sqrt{(a+3)^2+(b-1)^2}$

$\longrightarrow a^2+b^2+4b+4=a^2+6a+9+b^2-2b+1$

$\longrightarrow 4b+4=6a-2b+10$

$\longrightarrow b=a+1\quad\quad\dots(1)$

And,

$\longrightarrow\sqrt{(a-0)^2+(b+2)^2}=\sqrt{(a-1)^2+(b-3)^2}$

$\longrightarrow a^2+b^2+4b+4=a^2-2a+1+b^2-6b+9$

$\longrightarrow 4b+4=-2a-6b+10$

$\longrightarrow 10b=-2a+6$

From (1),

$\longrightarrow 10(a+1)=-2a+6$

$\longrightarrow 10a+10=-2a+6$

$\longrightarrow a=-\dfrac{1}{3}$

And then,

$\longrightarrow b=a+1$

$\longrightarrow b=-\dfrac{1}{3}+1$

$\longrightarrow b=\dfrac{2}{3}$

Hence the circumcenter of the triangle is,

$\longrightarrow\underline{\underline{(a,\ b)=\left(-\dfrac{1}{3},\ \dfrac{2}{3}\right)}}$