Here's a different approach for finding the two points P(x,y)

Since , draw a right triangle whose
opposite side is the numerator of 3/5, which is 3, and whose adjacent 
side is the denominator of 3/5, which is 5.  Then θ will be the 
angle with opposite side 3 and adjacent side 5:



Calculate the hypotenuse by the Pythagorean theorem:









We place the triangle on a graph so that the vertex is at the origin,
and draw a circle with center at the origin.  But this circle is not
the unit circle.



The unit circle has radius 1, but the above circle has radius √34, so we
divide everything by √34, to make the circle become the unit circle,
so we have the first value of P(x,y):



Also, since tangent is positive in QIII, we can also reflect everything
across the origin, and the angle is increased by 180°, and get another
point P(x,y).



Edwin