What is the Cartesian form of #( -2 , ( - 9pi)/2 ) #?

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Answer 1 · 2017-10-26T16:11:53

The Cartesian point is #(0,2)#

Given: #r = -2 and theta = (-9pi)/2#

For the x coordinate, use the equation:

#x = rcos(theta)#

#x = -2cos((-9pi)/2)#

#x = 0#

For the y coordinate, use the equation:

#y = rsin(theta)#

#y = -2sin((-9pi)/2)#

#y = 2#

The Cartesian point is #(0,2)#

Answer 2 · 2017-11-03T14:02:24

#( 0 , 2 )#

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It can be seen from the diagram that the x coordinate of P is #rcos(theta)# and the y coordinate of P is #rsin(theta)#

So:

#x=rcos(theta)#

#y=rsin(theta)#

From example:

#x=-2cos((-9pi)/2)= 0#

#y=-2sin((-9pi)/2)=2#

Cartesian coordinate:

#( 0 , 2 )#