Question

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

Coordinates Quadrant

$P\left(\begin{array}{c}\end{array},-\frac{2}{7}\right)$ IV

Short answer

The missing coordinate ofpoint $P$ is$x=\frac{3\sqrt{5}}{7}$.

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 $-\frac{2}{7}$ for $y$ into the equation $x^2+y^2=1$ and simplify for $x$.

localid="1648211091573" $$\begin{gathered} x^2+y^2=1 \\ {\left(x\right)}^2+{\left(-\frac{2}{7}\right)}^2=1 \\ {x}^2+\frac{4}{49}-\frac{4}{49}=1-\frac{4}{49} \\ {x}^2=\frac{45}{49} \\ x=\pm \frac{3\sqrt{5}}{7} \end{gathered}$$

Step 3. State the conclusion.

Since the point is in quadrant IV, therefore, $x$-coordinate of the point must be positive and hence, out of two calculated values of $x$, the only value that satisfies the condition is $x=\frac{3\sqrt{5}}{7}$.