Question

Find the missing coordinate of $P$, using the fact that $P$ lies on the unit circle in the given quadrant.

Coordinates Quadrant

$P\left(\begin{array}{c}\end{array},\frac{1}{3}\right)$ II

Short answer

The missing coordinate ofpoint $P$ is$x=-\frac{2\sqrt{2}}{3}$.

Step-by-step solution

Step 1. State the definition of the unit circle.

The unit circle is the circle of radius $1$ centered at the origin in the $xy$-plane. It is given by the equation $x^2+y^2=1$.

Step 2. Substitute the given points and simplify for x.

Since the point satisfies the equation of the unit circle, substitute $\frac{1}{3}$ for $y$ into the equation $x^2+y^2=1$and simplify for $x$.

role="math" localid="1648208414914" $$\begin{gathered} x^2+y^2=1 \\ {\left(x\right)}^2+{\left(\frac{1}{3}\right)}^2=1 \\ {x}^2+\frac{1}{9}-\frac{1}{9}=1-\frac{1}{9} \\ {x}^2=\frac{8}{9} \\ x=\pm \frac{2\sqrt{2}}{3} \end{gathered}$$

Step 3. State the conclusion.

Since the point is in quadrant II, therefore, $x$-coordinate of the point must be negative and hence, out of two calculated values of $x$, the only value that satisfies the condition is $x=-\frac{2\sqrt{2}}{3}$.