Rotating a line segment in a plane

Goofydeer

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Dec 1, 2011
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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
 
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.
 
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 \(\displaystyle 270^o\) CCW rotation is equivalent to a \(\displaystyle 90^o\) CW rotation.
And I know a trick for perpendicular lines.
Code:
            P"    |
      (-3,5)*     |
            :     |         P
           5:     |         *(5,3)
            :     |        3:
            :     |    5    :
      - - - + - - + - - + - + - -
              -3  |  3  :
                  |     :
                  |     :-5
                  |     :
                  |     *(3,-5)
                  |     P'
Suppose we have point \(\displaystyle P(5,3)\)

From the origin \(\displaystyle O\), we move 5 right and 3 up.

To find a point \(\displaystyle P'\) so that \(\displaystyle OP' \perp OP,\)
. . we move 3 right and 5 down ... to \(\displaystyle P'(3,-5)\)
. . . or 3 left and 5 up ... to \(\displaystyle P"(-3,5)\)

In either case, we [1] switch the x- and y-coordinates
. . . . . . . . . . and [2] change the sign of one of them.


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

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


Get it?
 
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