Question

Help! Unit Circles.
The point p is on the unit circle. Find p(x, y) from the given information.
The x coordinate of p is -sqroot6/4 and p lies below the x-axis.
p(x,y) =

Answer 1

Well, you can use circle formula.

Answer 2

The equation for a circle is: \[
(x-h)^2+(y-k)^2=r^2
\]In this case, \((h,k)\) the center is at \((0,0)\) and \(r\) the radius is \(1\).\[
x^2+y^2=1
\]Now we know that the point lies below the \(x\) axis, which means \(y<0\).
And we also know that \(x = -\sqrt{6}/4\)

Answer 3

Solving for \(y\) gives us: \[
y = \sqrt{1-x^2} = \sqrt{1-\left(\sqrt{6/4}\right)^2}
\]

Answer 4

So what would the final answer look like?
p= \[(-\sqrt{6}/4, y) \]
what is y exactly??

Answer 5

@wio

Answer 6

Oh, or is it p times (x,y) because it says look for p(x,y) ._.

Answer 7

Is the x coordinate of point p the expression \(\LARGE \frac{-\sqrt{6}}{4}\) ???

OR

Is the x coordinate of point p the expression \(\LARGE -\sqrt{\frac{6}{4}}\) ???

Answer 8

the first one

Answer 9

So if \[\Large x = \Large \frac{-\sqrt{6}}{4}\] then what is x^2 equal to?

Answer 10

\[(-\sqrt{6}/64^2\]?

Answer 11

ah! sorry that got messed up.

Answer 12

\[(-\sqrt{6}/4)^2\]

Answer 13

good, can you simplify that further?

Answer 14

\[\sqrt{36}/16\] would that be correct?

Answer 15

or..is the 36 still negative?

Answer 16

If you square \[\Large \sqrt{6}\], the square root will go away. This is because the square and square root undo each other

Answer 17

or you can think of it like \[\Large \sqrt{6*6}=\sqrt{36} = 6\]

Answer 18

either way, you'll have 6 in the numerator

Answer 19

Oh gosh. I knew that *facepalm. So then what do I do with the denominator?

Answer 20

you'll have 16 in the denominator since 4*4 = 16. So that is correct

Answer 21

x^2 = 6/16 = 3/8

make sure to reduce as much as possible

Answer 22

now you'll use this to find the value of y

the unit circle is the equation

x^2 + y^2 = 1

replace x^2 with 3/8 to get

(3/8) + y^2 = 1

now you need to solve for y

Answer 23

by putting y under a sq root?

Answer 24

how would you move that 3/8 over?

Answer 25

\[\sqrt{y}=1-(3/8)\]

Answer 26

or times both sides by 8?

Answer 27

You did it right. You subtract 3/8 from both sides.

Answer 28

1-(3/8) = ???

Answer 29

5/8!

Answer 30

so we know that

\[\Large y^2 = \frac{5}{8}\]

Answer 31

the ultimate goal is to find y itself

Answer 32

so you square root it.

Answer 33

?

Answer 34

\[\sqrt{y^2}=\sqrt{5}/8\]

Answer 35

the square root will apply to the entire number 5/8 and not just the 5

Answer 36

so

\[\Large y = \pm \sqrt{\frac{5}{8}}\]

we are told that "p lies below the x axis" so y must be negative, which means there is a negative sign outside the square root

\[\Large y = -\sqrt{\frac{5}{8}}\]

notice how the plus/minus turned to just a minus

Answer 37

Okay. So it's p\[(-\sqrt{6}/4,-\sqrt{5/8})\]

Answer 38

?

Answer 39

Correct, the point is

\[\Large \left(\frac{-\sqrt{6}}{4}, -\sqrt{\frac{5}{8}}\right)\]

Answer 40

So when it asks for p(x,y) I'm just meant to find the point and not multiply the p in..right? That would be a whole other thing xD

Answer 41

p(x,y) is basically the same as saying "p is the point in the form (x,y)"

it's unfortunate that it looks like "p times (x,y)" when that's not what it means

Answer 42

Ah, okay.
Sir Thompson, you have been absolutely amazing. Thank you so so much! you're a life saver.

Answer 43

Sorry i'm so slow xD tested into a math class so i'm a bit confused as to what all is going on.

Answer 44

you did fine, so no worries

Answer 45

Thanks again ^-^

Answer 46

you're welcome