Question

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

Answer 1

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

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