Answered

Suppose that (x, -8/17) is a point in quadrant IV lying on the unit circle.


Find x. Write the exact value, not a decimal approximation.


Can someone help me solve this problem?

Verified Answer :

LammettHash

Since the point lies on the unit circle, it satisfies

\(x^2+\left(-\dfrac8{17}\right)^2=x^2+\left(\dfrac8{17}\right)^2=1\)

Solving for \(x\) gives two possible solutions:

\(x^2=1-\left(\dfrac8{17}\right)^2\)
\(x=\pm\sqrt{1-\left(\dfrac8{17}\right)^2}\)

Given that \(x\) is in quadrant IV, you know that its value must be positive, so

\(x=\sqrt{1-\left(\dfrac8{17}\right)^2}=\sqrt{\dfrac{225}{289}}\)

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