finding vertices of a triangle after rotating [closed]

Let say that I have an equilateral triangle with vertices: (0,0) (200,346.4) (400,0). Say I want to rotate this triangle 30 degrees clockwise, how would I find the new vertices?

Edit1: Rotating about the center of the triangle, my apologies.

closed as off-topic by Chappers, user223391, Harish Chandra Rajpoot, graydad, J. W. PerrySep 26 '15 at 2:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Harish Chandra Rajpoot, graydad, J. W. Perry
If this question can be reworded to fit the rules in the help center, please edit the question.

• You should mention the vertex/point or the axis about which the triangle is rotated. – Harish Chandra Rajpoot Sep 25 '15 at 20:04
• Are you rotating about the origin? – David Quinn Sep 25 '15 at 20:04
• Yes rotating about the center of the triangle, my apologies – samuelk71 Sep 25 '15 at 20:04
• Are you looking for an analytic solution or a geometric one? – John Douma Sep 25 '15 at 20:06
• The actual vertices I am working on a website. -Thanks – samuelk71 Sep 25 '15 at 20:08

First subtract the coordinates of the centre, $(200,\frac{200\sqrt{3}}{3})$ from each vertex coordinate. Then multiply each new vertex by the rotation matrix for clockwise rotation by 30 degrees, i.e. $$\left(\begin{matrix}\cos(-30)&-\sin(-30)\\\sin(-30)&\cos(-30)\end{matrix}\right)$$ to obtain three new rotated vertices. Then add back the centre coordinates to each rotated vertex.

• Nice thats what I was looking for I was actually almost there haha. – samuelk71 Sep 25 '15 at 20:34

Here is how I would go about solving this;

• Imagine the three points of the triangle as points on a circle
"concentric" with the triangle.
• These points form three 120 degree arcs.
• The midpoints of these arcs are the vertices after a 60 degree rotation, and the quarterpoints of these arcs are the possible vertices after the two possible 30 degree rotations.

I am not at a point where I can do the calculations though, sorry. I would do this as a comment if I had that permission available to me on this stack. Hopefully this is able to point you in the right direction though!

• I was thinking something about that, but my roommate suggested something along the lines of a rotational matrix where you would use each line as a vector and rotate them individually. I'll look into both. – samuelk71 Sep 25 '15 at 20:16

Rotate the triangle about [b]what[/b] point?

I suspect you mean about the center of the triangle. If that is true, the first thing you need to do is determine that center. Fortunately, for a triangle, that is just the mean of the coordinates of the point, ((0+ 200+ 400)/3, (0+ 346.4/3))= (200, 115.5).

To determine what happens to the coordinates of a vertex when you wrote the triangle about that point: 1) Translate the whole triangle so that the center is moved to the orgin: Subtract the coordinates of the center from the coordinates of the vertex. (0, 0) becomes (-200, -115.5)

2) Rotate around the origin by angle $\theta[/tex]: The new x coordinate becomes [math]x'= x cos(\theta)- y sin(\theta)[/itex] and the new y coordinate [math]y'= x sin(\theta)+ y cos(\theta)$. With the values above and $\theta= 30$ degrees, $x'= (-200(0.8660)+ 115.5(.5)= -115.45$ and $y'= (-2oo)(.5)- 115.5(0.8660)= -200.023$. That gives (-115.45, -200.023)

3) Translate back to the center- add the coordinates of the center: (-115.45+ 200, -220.023+ 115.5)= (84.55, -104.523).