Question

Please solve the attachment below

Answer 1

Part 1)
`x=(+/-)(1)/(√(11))` ----->
`(x,(√(110))/(11))`

Part 2)
`x=(+/-)(6√(2))/(11)` ---->
`(x,(7)/(11))`

Part 3)
`x=(+/-)(4√(6))/(11)` ---->
`(x,(5)/(11))`

Part 4)
`x=(+/-)(2√(10))/(11)` --->
`(x,(9)/(11))`

Explanation:

we know that

In the unit circle

The coordinates of a point have the following rule

`x^(2) +y^(2) =r^(2)`

where

(x,y) are the coordinates of the point a r is the radius

but remember that in a unit circle the radius is equal to 1

so

`x^(2) +y^(2) =1`

`x^(2)=1-y^(2)`

`x=(+/-)\sqrt{1-y^(2)}`

Find the x-coordinate of each case

Part 1) we have the point

`(x,(√(110))/(11))`

so

The y-coordinate is

`y=(√(110))/(11)`

Find the value of the x-coordinate

substitute

`x=(+/-)\sqrt{1-y^(2)}`

`x=(+/-)\sqrt{1-((√(110))/(11))^(2)`

`x=(+/-)\sqrt{1-((110)/(121))`

`x=(+/-)\sqrt{(11)/(121))`

`x=(+/-)(√(11))/(11)`

`x=(+/-)(1)/(√(11))`

Part 2) we have the point

`(x,(7)/(11))`

so

The y-coordinate is

`y=(7)/(11)`

Find the value of the x-coordinate

substitute

`x=(+/-)\sqrt{1-y^(2)}`

`x=(+/-)\sqrt{1-((7)/(11))^(2)`

`x=(+/-)\sqrt{1-((49)/(121))`

`x=(+/-)\sqrt{(72)/(121))`

`x=(+/-)(6√(2))/(11)`

Part 3) we have the point

`(x,(5)/(11))`

so

The y-coordinate is

`y=(5)/(11)`

Find the value of the x-coordinate

substitute

`x=(+/-)\sqrt{1-y^(2)}`

`x=(+/-)\sqrt{1-((5)/(11))^(2)`

`x=(+/-)\sqrt{1-((25)/(121))`

`x=(+/-)\sqrt{(96)/(121))`

`x=(+/-)(4√(6))/(11)`

Part 4) we have the point

`(x,(9)/(11))`

so

The y-coordinate is

`y=(9)/(11)`

Find the value of the x-coordinate

substitute

`x=(+/-)\sqrt{1-y^(2)}`

`x=(+/-)\sqrt{1-((9)/(11))^(2)`

`x=(+/-)\sqrt{1-((81)/(121))`

`x=(+/-)\sqrt{(40)/(121))`

`x=(+/-)(2√(10))/(11)`