Find the coordinates of the points where the line $y = frac{1}{2}$ intersects the unit circle.

To find the intersection points of the line $y = \frac{1}{2}$ and the unit circle, we start with the equation of the unit circle, which is $x^2 + y^2 = 1$.

Substituting $y = \frac{1}{2}$ into the circle’s equation, we get:

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

Simplifying this, we obtain:

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

We then isolate $x^2$:

$x^2 = 1 – \frac{1}{4}$

Simplifying further:

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

Taking the square root of both sides, we get:

$x = \pm \frac{\sqrt{3}}{2}$

Thus, the intersection points are:

$(x, y) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$ and $(x, y) = \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right)$

To determine the intersection points of the line $y = frac{1}{2}$ with the unit circle, let’s begin with the unit circle’s equation $x^2 + y^2 = 1$.

Substitute $y = frac{1}{2}$:

$x^2 + left( frac{1}{2}
ight)^2 = 1$

Simplify to:

$x^2 + frac{1}{4} = 1$

Then:

$x^2 = 1 – frac{1}{4}$

Which gives:

$x^2 = frac{3}{4}$

Taking the positive and negative square roots, we find:

$x = pm frac{sqrt{3}}{2}$

The solution thus yields the points:

$(frac{sqrt{3}}{2}, frac{1}{2})$ and $( -frac{sqrt{3}}{2}, frac{1}{2})$

To find intersection points of the line $y = frac{1}{2}$ with the unit circle $x^2 + y^2 = 1$, substitute $y$:

$x^2 + left(frac{1}{2}
ight)^2 = 1$

Simplify:

$x^2 + frac{1}{4} = 1$

$x^2 = frac{3}{4}$

Hence:

$x = pm frac{sqrt{3}}{2}$

Coordinates are: $( frac{sqrt{3}}{2}, frac{1}{2})$ and $( -frac{sqrt{3}}{2}, frac{1}{2})$

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