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).
As with any "solve for ..." problem, you start by looking at the operations performed on the variable of interest. Here, when you evaluate this expression according to the order of operations, you
add b1
multiply by (h/2)
When you want to solve for b2, you undo these operations in reverse order. To undo multiplication by a fraction, you multiply by the inverse (reciprocal) of the fraction. To undo addition, you add the opposite.
Whatever you do must be done to both sides of the equation.
Here we go ...
... (2/h)A = b1 + b2 . . . . . we undid the multiply by h/2, by multiplying by 2/h
... (2/h)A - b1 = b2 . . . . . we undid the addition of b1 by adding the opposite of b1
Then your solution is
... b2 = 2A/h - b1
If you want to, you can combine these terms over a single denominator to get
