How do you convert #(theta)=-7pi/4# to rectangular form?

3 Answers
Shwetank Mauria
Feb 26, 2016

Equation in rectangular coordinates is #x+y=0#

Explanation:

To convert equation #theta=−(7pi)/4#, remember relation between polar and rectangular coordinates given by

#x=rcostheta# and #y=rsintheta# i.e. #r^2=x^2+y^2# and #theta=tan^(-1)(y/x)# or #tantheta=y/x#

Hence the equation #theta=−(7pi)/4# means

#tan (-(7pi)/4)=y/x#, but #tan(-(7pi)/4)=-1#

Hence equation in rectangular coordinates is #y/x=-1# i.e.

#x+y=0#

A. S. Adikesavan
Feb 26, 2016

The radial line (half line ) y = x, in the first quadrant..

Explanation:

In polar coordinates, #theta# = constant represents a radial line and is a half-line.
The other half in the opposite direction is governed by #theta# = the constant + #pi#.
Indeed, the given equation can be given directly as #theta# = #pi#/4.

P dilip_k
Feb 26, 2016

one more Explanation

Explanation:

#y=rsin(-7pi/4)#
#=>y=-rsin(7pi/4)# [since #sin(-theta)=-sintheta#]
#=>y=-rsin(2pi-pi/4)#
#=>y=rsin(pi/4)# [in 4th quadrant sin is negative]
#=>y=r/sqrt2#
again
#x=rcos(-7pi/4)# [since #cos(-theta)=costheta#]
#=>x=rcos(7pi/4)#
#=>x=rcos(2pi-pi/4)#
#=>x=rcos(pi/4)# [in 4th quadrant cos is positive]
#=>x=r/sqrt2#
hence
#x=y #

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