The point C(4,5) is rotated 270° counterclockwise around the origin. What are the coordinates of the resulting point, C'?
2 answers
Does it show a graph or anything so I can help?
Your rotation vector will be
cos270 , - sin270
sin 270, cos 270
so multiply this rotation vector by the vector
4
5
we know cos270 = 0 , sin 270= -1
so our result is the vector
0(4) - (-1)(5) = 5
-1(4) + (0)(5) = -4
so A(4,5) ----> A'(5,-4) after a rotation of 270°
Illustrate it with a sketch, we could just as well have rotated -90°,
that might have been easier to see in our sketch.
I can also verify my result using slopes
slope OA = (5-0)/(4-0) = 5/4
slope of OA' = (-4-0)/(5-0) = -4/5
so OA and OA' form a 90° angle, or an exterior angle of 270°
cos270 , - sin270
sin 270, cos 270
so multiply this rotation vector by the vector
4
5
we know cos270 = 0 , sin 270= -1
so our result is the vector
0(4) - (-1)(5) = 5
-1(4) + (0)(5) = -4
so A(4,5) ----> A'(5,-4) after a rotation of 270°
Illustrate it with a sketch, we could just as well have rotated -90°,
that might have been easier to see in our sketch.
I can also verify my result using slopes
slope OA = (5-0)/(4-0) = 5/4
slope of OA' = (-4-0)/(5-0) = -4/5
so OA and OA' form a 90° angle, or an exterior angle of 270°