Question

find the missing coordinate of p, using the fact that p lies on the unit circle in the given quadrant.
coordinates quadrant
$p(\frac{1}{4}, square)$ i
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Explanation:

Step1: Recall unit - circle equation

The equation of the unit circle is $x^{2}+y^{2}=1$. Given $x = \frac{1}{4}$, we substitute it into the equation.
$$(\frac{1}{4})^{2}+y^{2}=1$$

Step2: Solve for $y^{2}$

First, calculate $(\frac{1}{4})^{2}=\frac{1}{16}$. Then, we have $\frac{1}{16}+y^{2}=1$. Subtract $\frac{1}{16}$ from both sides: $y^{2}=1 - \frac{1}{16}=\frac{16 - 1}{16}=\frac{15}{16}$.

Step3: Solve for $y$

Take the square - root of both sides: $y=\pm\sqrt{\frac{15}{16}}=\pm\frac{\sqrt{15}}{4}$.

Step4: Determine the sign of $y$

Since the point $P$ lies in the first quadrant where both $x$ and $y$ are positive, we take the positive value of $y$. So $y = \frac{\sqrt{15}}{4}$.

Answer:

$\frac{\sqrt{15}}{4}$