To find the image point reflected over Y = X, we make the opposite moves from these points (reversed) on Line Y = X:įrom (2, 2) we move 1 unit to the right: X coordinate = 3įrom (3, 3) we move 1 unit down: Y coordinate = 2 This creates a right triangle with sides of 1 and 1 Point (2, 3) will be +1 unit above (2 ,2) and + 1 unit to the left of (3, 3) (2 ,2) and (3, 3) are both on line Y = X The original point of (2, 3) and the Image Point will always be equidistant from the Mirror Line over which the original point is reflected. Line Y = X is the Line that passes through the origin and creates a 45 degree angle with the X axis So the first rotation gives us point (2, 3)

Or, the explanation is too long winded, you can follow the Rule:įor 90 degree rotations clockwise about the Origin: (X, Y) ->becomes image point of (Y, -X) These 2 points and the origin will create a 90 degree angle about the Origin (both lines are diagonals of their respective rectangles) and connecting new image point A at (2, 3) connecting original point A at (-3 ,2) with the Origin Point A would now move to the upper right corner of this new rectangle that we pushed over -> this will be point (2, 3) This would give us a rectangle of 2 units along the positive X axis and 3 units along positive Y axis. Now imagine lifting the rectangle up and pushing it from quadrant 2 into quadrant 1 - i.e, a 90 degree rotation clockwise about the Origin If we join the horizontal and vertical lines to the Y and X axis, respectively, we end up with a rectangle that is 3 units along the Negative X Axis and 2 units along the Positive Y Axis Label (-3, 2) as point A and Origin as Point O Since we are rotating 90 degrees about the Origin (0, 0) -> make a rectangle with these 2 points The way I understood 90 degree rotations was by the “tipping the rectangle” method.

90 degrees clockwise rotation