Transformation-rotation

• Mar 12th 2010, 02:57 AM
mathhomework
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. :confused: 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.
• Mar 12th 2010, 07:46 AM
Soroban
Hello, mathhomework!

Quote:

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 . . .

• Mar 12th 2010, 02:22 PM
mathhomework
I can love math again!
Thank you!Thank you! Thank you! You are sooooo clear, I got it!(Clapping)
• Mar 12th 2010, 08:33 PM
mathhomework
Am I right?
Please clear this up for me. :confused:

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?(Worried)