Question

How do I find the points on the unit circle where the x-coordinate equals the negative of the y-coordinate?

Answer 1

To find the points on the unit circle where the x-coordinate equals the negative of the y-coordinate, we start by recalling the equation of the unit circle:

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

We want to find points where

\[
x = -y.
\]

To substitute this relationship into the unit circle equation, we can replace \(y\) with \(-x\):

\[
x^2 + (-x)^2 = 1.
\]

This simplifies to:

\[
x^2 + x^2 = 1,
\]
\[
2x^2 = 1.
\]

Dividing both sides by 2 gives:

\[
x^2 = \frac{1}{2}.
\]

Taking the square root of both sides, we find:

\[
x = \frac{1}{\sqrt{2}} \quad \text{or} \quad x = -\frac{1}{\sqrt{2}}.
\]

Now, substituting these values back into \(y = -x\):

1. If \(x = \frac{1}{\sqrt{2}}\), then \(y = -\frac{1}{\sqrt{2}}\).
This gives the point:
\[
\left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right).
\]

2. If \(x = -\frac{1}{\sqrt{2}}\), then \(y = \frac{1}{\sqrt{2}}\).
This gives the point:
\[
\left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right).
\]

Thus, the points on the unit circle where the x-coordinate equals the negative of the y-coordinate are:

\[
\left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) \quad \text{and} \quad \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right).
\]