Question

the point p=(6/7, y) lies on the unit circle below what is the value of y in simplest for

Answer 1

To find the value of \( y \) for the point \( P = \left( \frac{6}{7}, y \right) \) that lies on the unit circle, we start with the equation of the unit circle:

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

Substituting \( x = \frac{6}{7} \) into the equation, we have:

\[
\left( \frac{6}{7} \right)^2 + y^2 = 1
\]

Calculating \( \left( \frac{6}{7} \right)^2 \):

\[
\left( \frac{6}{7} \right)^2 = \frac{36}{49}
\]

Now we can substitute this back into the unit circle equation:

\[
\frac{36}{49} + y^2 = 1
\]

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

\[
y^2 = 1 - \frac{36}{49}
\]

To combine the terms on the right side, we need to express 1 as a fraction with a denominator of 49:

\[
1 = \frac{49}{49}
\]

Now substituting this in gives:

\[
y^2 = \frac{49}{49} - \frac{36}{49} = \frac{13}{49}
\]

Taking the square root to solve for \( y \):

\[
y = \pm \sqrt{\frac{13}{49}} = \pm \frac{\sqrt{13}}{7}
\]

Thus, the values of \( y \) are \( \frac{\sqrt{13}}{7} \) and \( -\frac{\sqrt{13}}{7} \).

Since you asked for the value of \( y \) in its simplest form, we can express it as:

\[
y = \pm \frac{\sqrt{13}}{7}
\]

If you need only one specific value (below the x-axis), then:

\[
y = -\frac{\sqrt{13}}{7}
\]