Find the missing coordinate of P, using the fact that P lies on the unit circle in the given quadrant.
a) P( ,-15/17) Quadrant IV
b) P(2/5, ) Quadrant I
the unit circle has a radius of 1 ... x^2 + y^2 = 1^2
a) x^2 = 1 - (225 / 289) = 64 / 289 ... x = 8 / 17
b) y^2 = 1 - (4 / 25) = 21 / 25
a) To find the missing coordinate of P in Quadrant IV on the unit circle, we can use the equation of the unit circle, which is x^2 + y^2 = 1. Since P is in Quadrant IV, the x-coordinate should be positive and the y-coordinate should be negative.
Let's use this information to find the missing coordinate. Given that P( , -15/17) is on the unit circle, we can square both sides of the equation x^2 + y^2 = 1 to get:
x^2 + (-15/17)^2 = 1
x^2 + 225/289 = 1
To solve for x, we can subtract 225/289 from both sides:
x^2 = 1 - 225/289
x^2 = 289/289 - 225/289
x^2 = 64/289
Taking the square root of both sides:
x = ± √(64/289)
Since P is in Quadrant IV, the x-coordinate should be positive. Therefore:
x = √(64/289)
Simplifying:
x = 8/17
So the missing coordinate of P in Quadrant IV is P(8/17, -15/17).
b) To find the missing coordinate of P in Quadrant I on the unit circle, we can again use the equation of the unit circle, which is x^2 + y^2 = 1. Since P is in Quadrant I, both the x-coordinate and y-coordinate should be positive.
Let's use this information to find the missing coordinate. Given that P(2/5, ) is on the unit circle, we can square both sides of the equation x^2 + y^2 = 1 to get:
(2/5)^2 + y^2 = 1
4/25 + y^2 = 1
To solve for y, we can subtract 4/25 from both sides:
y^2 = 1 - 4/25
y^2 = 25/25 - 4/25
y^2 = 21/25
Taking the square root of both sides:
y = ± √(21/25)
Since P is in Quadrant I, the y-coordinate should be positive. Therefore:
y = √(21/25)
Simplifying:
y = √21/5
So the missing coordinate of P in Quadrant I is P(2/5, √21/5).
To find the missing coordinate of point P on the unit circle, we need to use the fact that the point lies on the unit circle in the given quadrant.
a) P(-15/17, ) Quadrant IV:
In Quadrant IV, the x-coordinate is positive, and the y-coordinate is negative. Since the point lies on the unit circle, we know that the square of the x-coordinate plus the square of the y-coordinate equals 1.
Given that the x-coordinate is -15/17, we can find the missing y-coordinate as follows:
x^2 + y^2 = 1
(-15/17)^2 + y^2 = 1
225/289 + y^2 = 1
y^2 = 1 - 225/289
y^2 = (289/289) - (225/289)
y^2 = 64/289
y = ±√(64/289)
Since we are in quadrant IV and the y-coordinate is negative, the missing y-coordinate is -8/17.
Therefore, the missing coordinate of P is P(-15/17, -8/17).
b) P(2/5, ) Quadrant I:
In Quadrant I, both the x-coordinate and the y-coordinate are positive. Similar to the previous case, we use the equation x^2 + y^2 = 1 to find the missing y-coordinate.
Given that the x-coordinate is 2/5, we can find the missing y-coordinate as follows:
x^2 + y^2 = 1
(2/5)^2 + y^2 = 1
4/25 + y^2 = 1
y^2 = 1 - 4/25
y^2 = (25/25) - (4/25)
y^2 = 21/25
y = ±√(21/25)
Since we are in quadrant I and the y-coordinate is positive, the missing y-coordinate is √(21/25).
Therefore, the missing coordinate of P is P(2/5, √(21/25)).