My Math Forum triangle, rotate 45 degrees around the centroid

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 February 5th, 2017, 04:24 AM #1 Newbie   Joined: Jan 2017 From: ireland Posts: 8 Thanks: 0 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 07:10 AM.
 February 5th, 2017, 05:53 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,755 Thanks: 1405
 February 5th, 2017, 06:24 AM #3 Newbie   Joined: Jan 2017 From: ireland Posts: 8 Thanks: 0 Hello. thanks but I need the formula.
 February 5th, 2017, 06:39 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond See this. If you have any questions, post back.
February 5th, 2017, 07:01 AM   #5
Math Team

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Quote:
 Originally Posted by Gab 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 $(x',y')$ ...

$\begin{bmatrix} x'\\ y' \end{bmatrix}= \begin{bmatrix} \cos{\theta} &\sin{\theta} \\ -\sin{\theta}&\cos{\theta} \end{bmatrix} \cdot \begin{bmatrix} x-x_c\\ y-y_c \end{bmatrix}+ \begin{bmatrix} x_c\\ y_c \end{bmatrix}$

 February 5th, 2017, 08:52 AM #6 Newbie   Joined: Jan 2017 From: ireland Posts: 8 Thanks: 0 I don't know operations with matrices, can you do one and i can continue please?
 February 5th, 2017, 09:01 AM #7 Math Team   Joined: Jul 2011 From: Texas Posts: 2,755 Thanks: 1405 $\begin{bmatrix} x'\\ y' \end{bmatrix}= \begin{bmatrix} \cos{\theta} &\sin{\theta} \\ -\sin{\theta}&\cos{\theta} \end{bmatrix} \cdot \begin{bmatrix} x-x_c\\ y-y_c \end{bmatrix}+ \begin{bmatrix} x_c\\ y_c \end{bmatrix}$ 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 $x'$ ? Thanks from Gab
 February 5th, 2017, 09:17 AM #8 Newbie   Joined: Jan 2017 From: ireland Posts: 8 Thanks: 0 x′ = 2.45. thanks very much.
February 5th, 2017, 09:21 AM   #9
Math Team

Joined: Jul 2011
From: Texas

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Quote:
 Originally Posted by Gab x′ = 2.45. thanks very much.
I get $x' = 2.44$

 Tags centroid, degrees, rotate, triangle

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