Question #07602

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Answer 1 · 2017-03-31T22:10:27

The point is #(-3,-4)#

Given: #sec(theta) = -5/3#

We know that #sec(theta) = r/x#

We know that r cannot be negative, therefore, #x = -3# and #r = 5#

#r = sqrt(x^2+y^2)#

#5 = sqrt(-3^2+ y^2)#

#25 = 9+ y^2#

#y^2 = 16#

#y = +-4#

I quadrant 3, both x and y are negative so we choose the negative:

#y = -4#

The point is #(-3,-4)#