Question 278377: Give the polar coordinates: (-2, 2).
Found 2 solutions by nyc_function, Theo:
Answer by nyc_function(2741)   (Show Source):
You can put this solution on YOUR website! We need to convert to the form (r, theta).
Graph the given point on the xy-plane.
We drop a perpendicular line from (-2,2) to the x-axis forming a right triangle.
The value of x = -2 and the value of y = 2.
We now need to find the radius.
r = sqrt{(-2)^2 + (2)^2}
r = sqrt{8})
r = 2(sqrt{2})
We now need to find the value of theta.
We can use any trig function to find theta.
The tangent function = y/x. We know the values of x and y.
So, tan(theta) = 2/-2
Since theta is an angle, we need to use the tangent inverse key on the calculator.
We say that theta is -45 degrees in this example.
So, our final answer is (2(sqrt{2}), -45).
nycfunction@yahoo.com
Guidio
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! your cartesian coordinates are (-2,2)
x = -2
y = 2
you want to convert from the form (x,y) to the form (r,T) where r is the length of the line segment and T is the angle.
T can be expressed in degrees or radians.
Some formulas to help you.
r = sqrt(x^2 + y^2)
T = arcTan (y/x)
In your problem, r = sqrt(4 + 4) = sqrt(8) = 2.828427125
T = arcTan (2/-2) = arcTan (-1) = -45 degrees.
If you add 180 to it, then it can also be 135 degrees.
Since in the cartesian coordinate system, your point is in the second quadrant, then in the polar coordinate system it also has to be in the second quadrant.
-45 degrees is in the fourth quadrant.
180 - 45 = 135 degrees is in the second quadrant.
135 degrees is the angle that you want, so T = 135 degrees.
your point of (-2,2) in the cartesian coordinate system is equivalent to the point (2.828427125, 135) in the polar coordinate system.
Your polar coordinates are (2.828427125,135) where T is expressed in degrees.
A calculator that can help you convert between polar and cartesian coordinate system can be found in the following reference.
http://www.random-science-tools.com/maths/coordinate-converter.htm
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