Given P(1,-3); P'(-3,1) Q(3,-2);Q'(-2,3) R(3,-3);R'(-3,3) S(2,-4);S'(-4,2)
By observing the relationship between P and P', Q and Q',.... we note that (x,y)->(y,x) which corresponds to a single reflection about the line y=x. Alternatively, the same result may be obtained by first reflecting about the x-axis, then a positive (clockwise) rotation of 90 degrees, as follows: Sx(x,y)->(x,-y) [ reflection about x-axis ] R90(x,y)->(-y,x) [ positive rotation of 90 degrees ] combined or composite transformation R90. Sx (x,y)-> R90(x,-y) -> (y,x)
Similarly similar composite transformation may be obtained by a reflection about the y-axis, followed by a rotation of -90 (or 270) degrees, as follows: Sy(x,y)->(-x,y) R270(x,y)->(y,-x) => R270.Sy(x,y)->R270(-x,y)->(y,x)
So in summary, three ways have been presented to make the required transformation, two of which are composite transformations (sequence).
A point is zero-dimensional with respect to the covering dimension because every open cover of the space has a refinement consisting of a single open set.
If every line parallel to two lines intersects both regions in line segments of equal length, then the two regions have equal areas. In the case of your problem, every line parallel to the bases of the two parallelograms will intersect them in lines segments, each with a width of ℓ.