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$\frac{5}{13}$, and the $y$-coordinate is negative.

Short answer

The point$P\left(x,y\right)=\left(\frac{5}{13},-\frac{12}{13}\right)$.

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 $\frac{5}{13}$ for $x$ into the equation $x^2+y^2=1$ and simplify for $y$.

localid="1648215824286" $$\begin{gathered} x^2+y^2=1 \\ {\left(\frac{5}{13}\right)}^2+{\left(y\right)}^2=1 \\ \frac{25}{169}+y^2-\frac{25}{169}=1-\frac{25}{169} \\ {y}^2=\frac{144}{169} \\ y=\pm \frac{12}{13} \end{gathered}$$

Step 3. State the conclusion.

Since $y$-coordinate is given to be negative, therefore, out of two calculated values of $y$, the only value that satisfies the condition is $y=-\frac{12}{13}$.