Problem 666

If (x,y) is a point on the unit circle in quadrant IV and if x = \frac{\sqrt{2}}{2}, what is y?

 

Solution:-

 

If the point (x,y) is on the unit circle, then the point satisfies the equation of the unit circle

x^{2}+y^{2}=1

(\frac{\sqrt{2}}{2})^{2}+y^{2}=1

Evaluate (\frac{\sqrt{2}}{2})^{2}

 

\frac{1}{2}+ y^{2} = 1

 

Solve for y.

y^{2} =  \frac{1}{2}

 

Take the square root of both sides.

y = \pm \sqrt{\frac{1}{2}}

y = \pm  \frac{\sqrt{2}}{2}

 

In quadrant IV, the y-coordinate of the point is negative. Therefore, y = – \frac{\sqrt{2}}{2}

 

 

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