Thread: Transformation-rotation

Please help me. I don't know how to do this without drawing and tracing. That is very slow. This is middle school math. Please don't use sin and cos. Please expain this for me. Thank you.

Draw triangle LMN with vertices L(2, -1), M(6, -2) and N (4, 2). Find the coodinates of the vertices after a counterclockwise 90 degree rotation about the origin and about each of the point L, M and N.

Hello, mathhomework!

Draw $\displaystyle \Delta LMN$ with vertices: .$\displaystyle L(2, -1),\;M(6, -2),\;N (4, 2)$

Find the coodinates of the vertices after a counterclockwise 90° rotation:

(a) about the origin. . (b) about $\displaystyle L.$. . {c} about $\displaystyle M.$ . . (d) about $\displaystyle N.$

Luckily for both us, there are 90° rotations; there's a simple routine.


Suppose we have a point $\displaystyle P(3,2)$ relative to a point $\displaystyle O$

Code:

                    P
                    * (3,2)
                    ↑
                    ↑ 2
                    ↑
      O * → → → → → *
              3

Think of the slope of $\displaystyle OP\!:\;\;m_{_{OP}} \:=\:\frac{+2}{+3}$
. . That is, going from $\displaystyle O$ to $\displaystyle P$, we move: .$\displaystyle \frac{\text{ up 2}}{\text{right 3}} $

Let $\displaystyle P'$ be the new point after a 90° CCW rotation.

The slope of $\displaystyle PP'$ is perpendicular to $\displaystyle OP.$
. . Hence: .$\displaystyle m_{_{PP'}} \;=\;-\frac{3}{2} \quad\Rightarrow\quad \frac{-3}{2} \;=\;\frac{3}{-2}$

That is, going from $\displaystyle P$ to $\displaystyle P'$, we move: .$\displaystyle \frac{\text{down 3}}{\text{right 2}}\,\text{ or }\,\frac{\text{up 3}}{\text{left 2}} $


We know that $\displaystyle P'$ will be northwest of $\displaystyle P,$
. . so we use the second fraction: .$\displaystyle \frac{\text{up 3}}{\text{left 2}} $

From point $\displaystyle P$, we move: .left 2, up 3.

Code:

            P'
            * (1,5)
            ↑
            ↑
            ↑ 3
            ↑
            ↑   2
            * ← ← ← * (3,2)
                    :
                    : 2
                    :
      O * - - - - - *
              3

And we arrive at: .$\displaystyle P'(1,5)$



Do a few more of these and the steps will get smoother.
You'll come up with your own intuitive shortcuts.
Trust me on this . . .

Thank you!Thank you! Thank you! You are sooooo clear, I got it!

Please clear this up for me.

P rotates around O 90 degrees CCW, if I let O to be the origin, after the rotation P' will be (-2, 3) (I tried this and used tracing method.).

In your calculation, you didn't write the coordiantes of O. Does this mean O can be any point? But how come it does not work for (0, 0).

I looked at the tracing, the slopes of OP and OP' seem to perpendicular.

What did I do wrong?