Question

The point $P$ is on the unit circle. Find$P\left(x,y\right)$from the given information.

The $x$-coordinate of $P$ is$-\frac{\sqrt{2}}{3}$, and $P$ lies below the $x$-axis.

Short answer

The point$P\left(x,y\right)=\left(-\frac{\sqrt{2}}{3},-\frac{\sqrt{7}}{3}\right)$.

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 y.

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

role="math" localid="1648232818870" $$\begin{gathered} x^2+y^2=1 \\ {\left(-\frac{\sqrt{2}}{3}\right)}^2+{\left(y\right)}^2=1 \\ \frac{2}{9}+y^2-\frac{2}{9}=1-\frac{2}{9} \\ {y}^2=\frac{7}{9} \\ y=\pm \frac{\sqrt{7}}{3} \end{gathered}$$

Step 3. State the conclusion.

Since $P$ lies below the $x$-axis, so, $y$-coordinate must be negative, therefore, out of two calculated values of $y$, the only value that satisfies the condition is $y=-\frac{\sqrt{7}}{3}$.