Find the coordinates of a point on the negative unit circle

To find the coordinates of a point on the negative unit circle, we need to remember that the equation for a unit circle is $x^2 + y^2 = 1$. For a point on the negative unit circle, both x and y values will be negative.

Let’s take an example where $x = -\frac{1}{2}$. So,

$ x^2 + y^2 = 1 $

Substituting $x = -\frac{1}{2}$ into the equation, we get:

$ \left(-\frac{1}{2}\right)^2 + y^2 = 1 $

$ \frac{1}{4} + y^2 = 1 $

$ y^2 = 1 – \frac{1}{4} $

$ y^2 = \frac{3}{4} $

$ y = -\sqrt{\frac{3}{4}} $

$ y = -\frac{\sqrt{3}}{2} $

Therefore, the coordinates of the point are: $(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$.

Given the unit circle equation $x^2 + y^2 = 1$, we need to find coordinates on the negative unit circle where x and y are negative. Let’s use $x = -frac{3}{4}$. Then,

$ x^2 + y^2 = 1 $

Substituting $x = -frac{3}{4}$:

$ left(-frac{3}{4}
ight)^2 + y^2 = 1 $

$ frac{9}{16} + y^2 = 1 $

$ y^2 = 1 – frac{9}{16} $

$ y^2 = frac{7}{16} $

$ y = -sqrt{frac{7}{16}} $

$ y = -frac{sqrt{7}}{4} $

Therefore, the coordinates of the point are: $(-frac{3}{4}, -frac{sqrt{7}}{4})$.

Using the unit circle equation $x^2 + y^2 = 1$, let’s find a point on the negative unit circle.

Let $x = -frac{1}{sqrt{2}}$.

Then,

$ x^2 + y^2 = 1 $

$ left(-frac{1}{sqrt{2}}
ight)^2 + y^2 = 1 $

$ frac{1}{2} + y^2 = 1 $

$ y^2 = frac{1}{2} $

$ y = -frac{1}{sqrt{2}} $

Therefore, the coordinates are $(-frac{1}{sqrt{2}}, -frac{1}{sqrt{2}})$.

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