Question

which of the following could be points on the unit circle? a. (1/3,2/3 b. (4/3,4/5) c. (6/7,sq rt 13/7) d. (5/13,12/13)

Answer 1

freq. used Pythagorean triangles
3:4:5 and. 5:12:13

on unit circle, valid values for cos [x-axis] sin [y-axis] are
3/5, 4/5
5/13, 12/13

Answer 2

`((6)/(7),((√(13))/(7))` and
`((5)/(13) ,(12)/(13))` are the only two points that lies on unit circle.

Explanation:

The general equation for the circle is
`(x-h)^2+(y-k)^2=r^2` where (h,k) is the point of center.

So, For a circle with center at (0,0) and radius 1 unit. equation becomes,

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

Now, from the given points are on unit circle or not , we just have to put the values for x and y and check whether they satisfy the equation of circle.

1)
`((1)/(3) ,(2)/(3))`

Substitute for x and y in Left side of (1), we get,

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

`\Rightarrow ((1)/(3))^2+((2)/(3))^2`

`\Rightarrow ((1)/(9))+((4)/(9))`

`\Rightarrow ((1+4)/(9))=(5)/(9) \\eq 1`

Thus,
`((1)/(3) ,(2)/(3))` point do not lies on unit circle.

2)
`((4)/(3) ,(4)/(5))`

Substitute for x and y in Left side of (1), we get,

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

`\Rightarrow ((4)/(3))^2+((4)/(5))^2`

`\Rightarrow ((16)/(9))+((16)/(25))`

`\Rightarrow ((544)/(225)) \\eq 1`

Thus,
`((4)/(3) ,(4)/(5))` point do not lies on unit circle.

3)
`((6)/(7) ,(√(13))/(7))`

Substitute for x and y in Left side of (1), we get,

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

`\Rightarrow ((6)/(7))^2+((√(13))/(7))^2`

`\Rightarrow ((36)/(49))+((13)/(49))`

`\Rightarrow ((36+13)/(49))=(49)/(49)=1`

Thus,
`((6)/(7) ,(√(13))/(7))` point lies on unit circle.

4)
`((5)/(13) ,(12)/(13))`

Substitute for x and y in Left side of (1), we get,

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

`\Rightarrow ((5)/(13))^2+((12)/(13))^2`

`\Rightarrow ((25)/(169))+((144)/(169))`

`\Rightarrow ((25+144)/(169))=(169)/(169)=1`

Thus,
`((5)/(13) ,(12)/(13))` point lies on unit circle.

Thus, only two points lies on unit circle.

`((6)/(7) ,(√(13))/(7))` and
`((5)/(13) ,(12)/(13))` are the only two points that lies on unit circle.