Question

Find the image of R(5,6) under 90° clockwise rotation about a point (3,2)

Answer 1

The image is R'(7,0). So how do we get that?

There are various ways to find the image.

Let O be the point (3,2). One way is to find the line that passes through O and R:

Gradient of OR=(6-2)/(5-3)=4/2=2.

Now plug in the coordinates of O: y-2=2(x-3), y=2x-6+2=2x-4. So the equation of OR is y=2x-4.

OR' has the gradient -½ because it's perpendicular to OR and it also passes through O:

y-2=-½(x-3)=-x/2+3/2, y=-x/2+3/2+2=-x/2+7/2.

So R' lies somewhere on this line, but where?

Since OR and OR' have the same length they could be the radii of the circle centre O and radius²=(6-2)²+(5-3)²=4²+2²=20.

The equation of the circle is (x-3)²+(y-2)²=20 and the circle meets the line OR' in two places given by substituting for y or x in the equation of the circle. But it's actually easier to use y-2=-½(x-3), because we already have the expression x-3=-2(y-2), so we can substitute for x-3:

4(y-2)²+(y-2)²=20, 5(y-2)²=20, (y-2)²=4, y-2=±2, so y=2-2=0 or y=2+2=4. The rotation is clockwise so y=0 is the y-coord of R'.

The x-coord can by found by using the equation of OR' and plugging in y=0:

-2=-½(x-3), 4=x-3, so x=7, making the point R'(7,0).

Another way is to look at the geometry by plotting the points and observing what happens to the vector OR as it rotates into its perpendicular vector OR'.

ONR is a right triangle where N=90°=N'ÔN and RN=6-2=4, ON=5-3=2. After rotation, N(5,2) moves to N'(3,0), and ON'=ON=2. R'N'=RN=4. x-coord of R'=x-coord of O + R'N'=3+4=7. y-coord of R'=y-coord of O - ON'=2-2=0. Hence R'(7,0). Note that triangles ORN and OR'N' are congruent.