Question

A special right triangle drawn in Quadrant 1 intersects the unit circle at point (x, y) where y = 2√2
. What is the x-coordinate? (1 point)

Answer 1

To find the x-coordinate of the point \((x, y)\) where \(y = 2\sqrt{2}\) on the unit circle, we start from the equation of the unit circle, which is given by:

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

Substituting \(y = 2\sqrt{2}\) into the equation:

\[
x^2 + (2\sqrt{2})^2 = 1
\]

Calculating \((2\sqrt{2})^2\):

\[
(2\sqrt{2})^2 = 4 \cdot 2 = 8
\]

Now substitute this value back into the equation:

\[
x^2 + 8 = 1
\]

To isolate \(x^2\):

\[
x^2 = 1 - 8
\]
\[
x^2 = -7
\]

Since \(x^2 = -7\) does not yield a real value for \(x\), it indicates that the point \((x, y) = (x, 2\sqrt{2})\) cannot lie on the unit circle.

Therefore, there is no x-coordinate for the given y-coordinate of \(2\sqrt{2}\) on the unit circle. The coordinates do not intersect the unit circle.

**Final Answer:** No valid x-coordinate exists (y = 2√2 does not lie on the unit circle).