What equation would you use to show that the point ((square root of 11)/(6), (5/6))is on the unit circle?
show that
(√11/6)^2 + (5/6)^2 = 1 , (it is)
show that the point is on the unit circle is (√11/6 , 5/6)
To show that a point is on the unit circle, we can use the Pythagorean theorem equation, which relates the coordinates of a point on the unit circle. The equation is:
x^2 + y^2 = 1
In this case, we need to substitute the given coordinates of the point ((√11)/6, 5/6) into the equation and check if it satisfies the equation.
Substituting the x-coordinate of the point, (√11)/6, and the y-coordinate, 5/6, into the equation, we get:
((√11)/6)^2 + (5/6)^2 = 1
Simplifying the equation, we have:
(11/36) + (25/36) = 1
Combining the fractions on the left side of the equation, we get:
(11 + 25) / 36 = 1
36/36 = 1
Therefore, the equation is true, indicating that the point ((√11)/6, 5/6) lies on the unit circle.