Given one set of coordinates, find another set that is 120 degrees away.

**justinjrichards** · April 19th 2012, 10:06 AM

given the set of coordinates (-1035,-410) how would i go about finding another set of coordinates that is 120 degrees away?

**Plato** · April 19th 2012, 10:14 AM

Originally Posted by justinjrichards

given the set of coordinates (-1035,-410) how would i go about finding another set of coordinates that is 120 degrees away?

What does "120 degrees away" mean to you?
Please explain.

**justinjrichards** · April 19th 2012, 12:34 PM

its been a while since i took trig. so my terminology may be incorrect, but if the given coordinates are on the circumference of the circle then there could be two more sets of coordinates on the circle that are 120 degrees apart. one set clockwise 120 degrees and another set counter clockwise 120 degrees. the center of the circle would be at (0,0) i believe.

**Plato** · April 19th 2012, 12:59 PM

Originally Posted by justinjrichards

its been a while since i took trig. so my terminology may be incorrect, but if the given coordinates are on the circumference of the circle then there could be two more sets of coordinates on the circle that are 120 degrees apart. one set clockwise 120 degrees and another set counter clockwise 120 degrees. the center of the circle would be at (0,0) i believe.

If $(-1035,-410)$ is on a circle centered at $(0,0)$ then the radius is $1113.25$ .
You posted this in a trigonometry forum. The answer I will give may not look correct in trig terms.

This is done with a rotation matrix, a transformation.
$\left( {x\cos (\theta ) - y\sin (\theta ),x\sin (\theta ) + y\cos (\theta )} \right)$ is the resulting point of rotating the point $(x,y)$ about $(0,0)$ through an angle of $\theta$ .

**skeeter** · April 19th 2012, 03:25 PM

Originally Posted by justinjrichards

given the set of coordinates (-1035,-410) how would i go about finding another set of coordinates that is 120 degrees away?

let $(x_1,y_1)$ be the starting set of coordinates.

$r = \sqrt{x_1^2 + y_1^2}$

since $x_1$ and $y_1$ are both negative, the angle of the ray that passes thru $(x_1,y_1)$ is $\theta = \arctan\left(\frac{y_1}{x_1}\right) + 180^\circ$

$120^\circ$ "away" could be $(\theta + 120^\circ)$ or $(\theta - 120^\circ)$

new coordinates could be $(r\cos(\theta + 120^\circ),r\sin(\theta + 120^\circ))$ or $(r\cos(\theta - 120^\circ),r\sin(\theta - 120^\circ))$

**Plato** · April 19th 2012, 03:47 PM

With regard to reply #5 there is a way to simply cut to the chase directly.
Using $\left( {x\cos (\theta ) - y\sin (\theta ),x\sin (\theta ) + y\cos (\theta )} \right)$ let $x=-1035~\&~y=-410$ then take two cases $\theta =\pm 120^o$ .

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