Geometry- How do I rotate an image about a point other than (0,0)?

Question

This is the question: Given PQ with P(–2, 2) and Q(4, 4) is rotated 90 degrees counterclockwise with a center of rotation at A(2, –1), graph and list the coordinates of P'Q' .

What are the new coordinates, and how can I find them?

Answer 1

For a 90º counter-clockwise rotation from any point, what will happen is that the segment connecting one endpoint of the given segment to the point of rotation and the segment connecting the image point of that endpoint to the point of rotation will be perpendicular to each other, meaning their slopes must be opposite reciprocals. So, in this case, segment PA will be perpendicular to segment P'A, and segment QA will be perpendicular to segment Q'A.

So let's begin with PA. Using slope formula, the slope of PA is ( -1-2)/(2+2) or -3/4. So this means that segment P'A will have to have a slope of 4/3. Now, starting at point A, our point of rotation, we will use this slope to move to the image point, P', by the rise and run values of the slope. ** Note: we will work with our slope of 4/3 as -4/-3 to guarantee that we are moving in a counterclockwise direction. So we drop 4 units for the y-coordinate of A and we move left 3 units for the x-coordinate of A, and this brings us to P' at ( -1, -5 ).

We will follow the same procedure for rotating Q. The slope of QA is 5/2, so this means the slope of segment Q'A will have to be -2/5 or 2/-5. So, from A again, we will now move 2 units up for the y-coordinate and move left 5 units for the x-coordinate, and this brings up to Q' at ( -1, 6). ** Note: If the original segment and point of rotation are plotted on the coordinate plane, the graphs of P' and Q' will lend visual support to the 90 degree rotation.