Given the coordinates of the following locations
[tex]\begin{gathered} Q(4,-2) \\ R(2,-4) \\ S(0,2) \end{gathered}[/tex]
From the question we have that the distance of the point are equal so we will have;
[tex](x-4)^2+(y+2)^2=(x-2)^2+(y+4)^2=(x-0)^2+(y-2)^2[/tex]
Solving equation 1 and 3 simultaneously we will have
[tex]\begin{gathered} x^2-8x+16+y^2+4y+4=x^2+y^2-4y+4_{} \\ -8x+8y=-16 \\ \text{Divide through by 8} \\ -x+y=-2 \\ x=y+2 \end{gathered}[/tex]
Solving equation 2 and 3 simultaneously we will have
[tex]\begin{gathered} x^2-4x+4+y^2+8y+16=x^2+y^2-4y+4 \\ -4x+12y=-16 \\ \text{Divide through by 4} \\ -x+3y=-4 \\ x=3y+4 \end{gathered}[/tex]
Thus , to solve for y we have;
[tex]\begin{gathered} y+2=3y+4 \\ 2-4=3y-y \\ -2=2y \\ y=\frac{-2}{2}=-1 \end{gathered}[/tex]
Substitute y to find x
[tex]\begin{gathered} x=y+2 \\ x=-1+2=1 \end{gathered}[/tex]
Hence the coordinates of the center of the merry-go-round is ( 1, - 1)
The second option is the correct option
