Question

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

Coordinates Quadrant

$P\left(-\frac{3}{5},\begin{array}{c}\end{array}\right)$ III

Short answer

The missing coordinate ofpoint $P$ is$y=-\frac{4}{5}$.

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 $-\frac{3}{5}$ for $x$ into the equation $x^2+y^2=1$ and simplify for $y$.

role="math" localid="1648207135448" $$\begin{gathered} x^2+y^2=1 \\ {\left(-\frac{3}{5}\right)}^2+{\left(y\right)}^2=1 \\ \frac{9}{25}+y^2-\frac{9}{25}=1-\frac{9}{25} \\ {y}^2=\frac{16}{25} \\ y=\pm \frac{4}{5} \end{gathered}$$

Step 3. State the conclusion.

Since the point is in quadrant III, therefore, $y$-coordinate of the point must be negative and hence, out of two calculated values of $y$, the only value that satisfies the condition is $y=-\frac{4}{5}$.