A special right triangle drawn in Quadrant I intersects the unit circle at point (x,y)
where y=2√2
. What is the value of x in point (x,y)
? Use the equation of the unit circle to determine the missing coordinate.(1 point)
The equation of the unit circle is x^2 + y^2 = 1.
Since we are given that y = 2√2, we can substitute this value into the equation of the unit circle:
x^2 + (2√2)^2 = 1
x^2 + 8 = 1
x^2 = -7
Since x^2 is negative, there is no real solution to this equation. Therefore, there is no value of x in the point (x,y) when y = 2√2.