Co-ordinates of A, B, c and D fron graph are A(3, 5), B(7, 9), C(11, 5) and D(7, 1).
To find the shape of `squareABCD` :
`" "AB^(2)=(7-3)^(2)+(9-5)^(2)=4^(2)+4^(2)=32`
`rArr" "AB=4sqrt(2)` units
`" "BC^(2)=(11-7)^(2)+(5-9)^(2)=(4)^(2)+(-4)^(2)=32`
`rArr" "BC=4sqrt(2)` units
`" "CD^(2)=(7-11)^(2)+(1-5)^(2)=(-4)^(2)+(-4)^(2)=32`
`rArr" "CD=4sqrt(2)` units
`" "DA^(2)=(7-3)^(2)+(1-5)^(2)=4^(2)+(-4)^(2)=32`
`rArr" "DA=sqrt(4^(2)(1+1))=4sqrt(2)` units
`therefore" "AB=BC=CD=DA=4sqrt(2)` units
So, ABCD will be either square or rhombus.
Now, `" ""Diagonal "AC=sqrt((11-3)^(2)+(5-5)^(2))`
`rArr" "AC=sqrt((8)^(2)+(0)^(2))" "rArrAC=8` units
and `" ""diagonal "BD=sqrt((7-7)^(2)+(1-9)^(2))=sqrt((0)^(2)+(-8)^(2))=sqrt(8^(2))`
`rArr" "BD=8` units
`therefore " ""Diagonal "AC="Diagonal "BD`
So, given quadrilateral ABCD is a squre. The point which is equidistant from point A, B, C, D of a square ABCD will be at the intersecting point of diagonals and diagonals bisects each other.
Hence, the required point O equidistant from A, B, C, D is mid-point of any diagonal
`=((7+7)/(2), (9+1)/(2))=((14)/(2), (10)/(2))=(7, 5)`.
Hence, the required point is (7, 5).