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(A) First,find the vertices of the triangle by solving the equations of the sides in pairs:$1$. Intersection of $x - y + 1 = 0$ and $y - 1 = 0$: Subst

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$(2, 1)$

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(A) First,find the vertices of the triangle by solving the equations of the sides in pairs:$1$. Intersection of $x - y + 1 = 0$ and $y - 1 = 0$: Substituting $y = 1$ into $x - y + 1 = 0$ gives $x - 1 + 1 = 0$,so $x = 0$. Vertex $A = (0, 1)$.$2$. Intersection of $x + y - 5 = 0$ and $y - 1 = 0$: Substituting $y = 1$ into $x + y - 5 = 0$ gives $x + 1 - 5 = 0$,so $x = 4$. Vertex $B = (4, 1)$.$3$. Intersection of $x + y - 5 = 0$ and $x - y + 1 = 0$: Adding the equations gives $2x - 4 = 0$,so $x = 2$. Substituting $x = 2$ into $x + y - 5 = 0$ gives $2 + y - 5 = 0$,so $y = 3$. Vertex $C = (2, 3)$.Now,check the slopes of the sides:Slope of $AB$ $(y=1)$ is $0$.Slope of $AC$ $(x-y+1=0)$ is $1$.Slope of $BC$ $(x+y-5=0)$ is $-1$.Since the product of slopes of $AC$ and $BC$ is $(1) 	imes (-1) = -1$,the triangle is right-angled at $C(2, 3)$.For a right-angled triangle,the circumcentre is the midpoint of the hypotenuse $AB$.Midpoint of $AB = (\frac{0+4}{2}, \frac{1+1}{2}) = (2, 1)$.

The equations of the sides of a triangle are $x + y - 5 = 0$,$x - y + 1 = 0$,and $y - 1 = 0$. Find the coordinates of the circumcentre.

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