My Math Forum triangle, rotate 45 degrees around the centroid
 User Name Remember Me? Password

 New Users Post up here and introduce yourself!

 February 5th, 2017, 05:24 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 08:10 AM.
 February 5th, 2017, 06:53 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,568 Thanks: 1271
 February 5th, 2017, 07:24 AM #3 Newbie   Joined: Jan 2017 From: ireland Posts: 8 Thanks: 0 Hello. thanks but I need the formula.
 February 5th, 2017, 07:39 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,512 Thanks: 908 Math Focus: Elementary mathematics and beyond See this. If you have any questions, post back.
February 5th, 2017, 08:01 AM   #5
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,568
Thanks: 1271

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, 09:52 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, 10:01 AM #7 Math Team   Joined: Jul 2011 From: Texas Posts: 2,568 Thanks: 1271 $\begin{bmatrix} x'\\ y' \end{bmatrix}= \begin{bmatrix} \cos{\theta} &\sin{\theta} \\ -\sin{\theta}&\cos{\theta} \end{bmatrix} \cdot \begin{bmatrix} x-x_c\\ y-y_c \end{bmatrix}+ \begin{bmatrix} x_c\\ y_c \end{bmatrix}$ 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, 10:17 AM #8 Newbie   Joined: Jan 2017 From: ireland Posts: 8 Thanks: 0 x′ = 2.45. thanks very much.
February 5th, 2017, 10:21 AM   #9
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,568
Thanks: 1271

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

 Tags centroid, degrees, rotate, triangle

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Grozny Geometry 9 April 3rd, 2016 09:56 AM budski52 Trigonometry 7 February 23rd, 2015 01:16 PM DecoratorFawn82 Algebra 6 September 18th, 2014 08:48 AM mathmaniac Algebra 2 February 14th, 2013 06:04 PM grogmachine Algebra 3 August 5th, 2011 01:16 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top