How do you solve this pre-calculus problem?

Question

If the point P(13/14 , y) is on the unit circle in quadrant IV, then y=


anyone know how to even start to solve for this?

Answer 1

All points (x,y) on the unit circle satisfy the quation x^2 + y^2 =1.

In your case, x=13/14 so (13/14)^2 +y^2 =1.
Then make y the subject y= sqrt (1- (13/14)^2).
You'll get two answers for the square root. (Hint: only one of them is in quadrant IV!)

(It's the negative one - Quadrant IV has positive x and negative y)

Answer 2

The unit circle is a circle of radius 1 centred on the origin.

Quadrant IV is the quadrant to the right of the y axis below the x axis.

The x value is 13/14 the y value will be negative.

As it is a unit circle with radius 1 ---- x^2 + y^2 = 1

Hence y^2 = 1 - x^2

and y = sqrt (1 - (13/14)^2)

y = sqrt(1 - (169/196))

y = sqrt(27/196) = +/- 0.37

Take the negative value y = -0.37

Answer 3

Draw x and y axes.
From origin, P is point 13/14 to the right and then y down.
OP is then hypotenuse of right angled triangle of sides length 13/14 and y
OP is of length 1
1² = (13/14)² + y²
y² = 1² - (13/14)² = (1 - 13/14) (1 + 13/14)
y² = 1/14 x 27/14
y = √27 / 14 = 3√3 / 14 = 0.37
Check
(13/14)² + 0.37² = 1 as required radius
P is then point (13/14, - 0.37)

Answer 4

Use the Pythagorean Theorem.

x² + y² = 1
y² = 1 - x²
y² = 1 - (13/14)² = 1 - 169/196 = 27/196
y = -√(27/196) = -(3√3)/14 (y is negative in the fourth quadrant)
y ≈ -0.3711537

Answer 5

Unless I'm missing something, its just a matter of geometry. A right angled triangle. Hypotenuse = 1. One side length 13/14.
Use Pythagoras: 1^2 - (13/14)^2 = y^2
Some manipulation: 27/196 = y^2
Y = +-0.37
As P is in Quadrant IV, use y = - 0.37

Answer 6

simple. use pythagoras theorem. ( 13/14) ^2 + (y)^2 = (1)^2

y = (27/ 169) ^ 1/2
y= (0.16)^1/2
y= 0.4