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) is this a multiply choice? these are the options
√2/2
√3/2
1

1 answer

Given that the point on the unit circle is \((x, y)\) where \(y = \frac{\sqrt{2}}{2}\), we know that the equation of the unit circle is \(x^2 + y^2 = 1\).

Substituting \(y\) into the equation:

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

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

\[ \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \]

Now substituting this value back into the unit circle equation:

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

Subtract \(\frac{1}{2}\) from both sides:

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

Taking the square root of both sides gives:

\[ x = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2} \]

Since we are in Quadrant 1, where \(x\) is positive, we take the positive root:

Thus, the x-coordinate is:

\[ x = \frac{\sqrt{2}}{2} \]

Therefore, the correct answer is:

\(\sqrt{2}/2\)

Answer: \(\sqrt{2}/2\)