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February 5th, 2017, 05:24 AM   #1
Gab
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triangle, rotate 45 degrees around the centroid

I have a triangle and its center and I need to rotate it 45 degrees around its center in clockwise direction. How can I do it please? The coordinates are
A(5,20), B(5,30), C(15,25), centroid is (8.33, 25)

Last edited by skipjack; February 5th, 2017 at 08:10 AM.
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February 5th, 2017, 06:53 AM   #2
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February 5th, 2017, 07:24 AM   #3
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Hello. thanks but I need the formula.
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February 5th, 2017, 07:39 AM   #4
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February 5th, 2017, 08:01 AM   #5
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Quote:
Originally Posted by Gab View Post
Hello. thanks but I need the formula.
What do you know about operations with matrices?

For a point $(x,y)$ to be rotated about the point $(x_c,y_c)$ by an angle, $\theta$, clockwise, resulting in a rotated image point $(x',y')$ ...

$\begin{bmatrix}
x'\\ y'
\end{bmatrix}=
\begin{bmatrix}
\cos{\theta} &\sin{\theta} \\
-\sin{\theta}&\cos{\theta}
\end{bmatrix} \cdot
\begin{bmatrix}
x-x_c\\ y-y_c
\end{bmatrix}+
\begin{bmatrix}
x_c\\ y_c
\end{bmatrix}$
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February 5th, 2017, 09:52 AM   #6
Gab
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I don't know operations with matrices, can you do one and i can continue please?
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February 5th, 2017, 10:01 AM   #7
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$\begin{bmatrix}
x'\\ y'
\end{bmatrix}=
\begin{bmatrix}
\cos{\theta} &\sin{\theta} \\
-\sin{\theta}&\cos{\theta}
\end{bmatrix} \cdot
\begin{bmatrix}
x-x_c\\ y-y_c
\end{bmatrix}+
\begin{bmatrix}
x_c\\ y_c
\end{bmatrix}$

the above matrix translates into two equations ...

$x' = \cos{\theta}(x-x_c) + \sin{\theta}(y - y_c) + x_c$

$y' = -\sin{\theta}(x-x_c) + \cos{\theta}(y - y_c) + y_c$

using the point $(x,y) = (5,20)$ and $(x_c,y_c) = (8.33,25)$ in the first equation to find the rotated x-value, $x'$ ...

$x' = \cos(45^\circ)(5-8.33) + \sin(45^\circ)(20 - 25) + 8.33$

get out your calculator ... what do you get for $x'$ ?
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February 5th, 2017, 10:17 AM   #8
Gab
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x′ = 2.45. thanks very much.
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February 5th, 2017, 10:21 AM   #9
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Quote:
Originally Posted by Gab View Post
x′ = 2.45. thanks very much.
I get $x' = 2.44$
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