Goofydeer

12-01-2011, 09:24 AM

Plot the points E(1, −2), and F(4, −3) on a graph, and draw the line segment that connects them. Now rotate line segment EF by 270° counterclockwise about (0, 0).

I plotted the new points as E' (-2,-1) and F'(-3, -4), but this does not look right. Please help

tkhunny

12-01-2011, 09:54 AM

Why do you care how it looks? Maybe you're not a great artist!

IS it correct? This is the question.

You can also comprehend such problems by rotating the axis in the opposite direction.

soroban

12-01-2011, 11:29 AM

Hello, Goofydeer!

Plot the points E(1, −2), and F(4, −3) on a graph, and draw the line segment that connects them.

Now rotate line segment EF by 270° counterclockwise about (0, 0).

I plotted the new points as E' (-2,-1) and F'(-3, -4), but this does not look right.Why do you doubt your work? .Your answers are correct!

I almost "eyeballed" the problem.

A 270^o CCW rotation is equivalent to a 90^o CW rotation.

And I know a trick for perpendicular lines.

P" |

(-3,5)* |

: | P

5: | *(5,3)

: | 3:

: | 5 :

- - - + - - + - - + - + - -

-3 | 3 :

| :

| :-5

| :

| *(3,-5)

| P'

Suppose we have point P(5,3)

From the origin O, we move 5 right and 3 up.

To find a point P' so that OP' \perp OP,

. . we move 3 right and 5 down ... to P'(3,-5)

. . . or 3 left and 5 up ... to P"(-3,5)

In either case, we [1] switch the x- and y-coordinates

. . . . . . . . . . and [2] change the sign of one of them.

So 90^o from (1,\text{-}2) is at: .(2,1) or (\text{-}2,\text{-}1)

and 90^o from (4,\text{-}3) is at: .(3,4) or (\text{-}3,\text{-}4)

Get it?

Powered by vBulletin® Version 4.2.2 Copyright © 2016 vBulletin Solutions, Inc. All rights reserved.