Question

the point p=(- 1/7, y) lies on the unit circle below what is the value of y in simplest for

Answer 1

To determine the value of \( y \) for the point \( P = \left(-\frac{1}{7}, y\right) \) that lies on the unit circle, we can use the equation of the unit circle, which is given by:

\[
x^2 + y^2 = 1
\]

In your case, \( x = -\frac{1}{7} \). Substituting this value into the unit circle equation:

\[
\left(-\frac{1}{7}\right)^2 + y^2 = 1
\]

Calculating \(\left(-\frac{1}{7}\right)^2\):

\[
\left(-\frac{1}{7}\right)^2 = \frac{1}{49}
\]

Now substituting this back into the equation:

\[
\frac{1}{49} + y^2 = 1
\]

Next, we will isolate \( y^2 \):

\[
y^2 = 1 - \frac{1}{49}
\]

To subtract \(\frac{1}{49}\) from 1, we can write 1 as \(\frac{49}{49}\):

\[
y^2 = \frac{49}{49} - \frac{1}{49} = \frac{48}{49}
\]

Now, taking the square root to find \( y \):

\[
y = \pm \sqrt{\frac{48}{49}}
\]

We can simplify \(\sqrt{\frac{48}{49}}\):

\[
y = \pm \frac{\sqrt{48}}{7}
\]

The square root of 48 can be simplified as follows:

\[
\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}
\]

Thus, we have:

\[
y = \pm \frac{4\sqrt{3}}{7}
\]

Since you mentioned "below," we can infer you want the negative value:

\[
y = -\frac{4\sqrt{3}}{7}
\]

Therefore, the value of \( y \) in simplest form is:

\[
\boxed{-\frac{4\sqrt{3}}{7}}
\]