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)

Respuesta :

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:

[tex](\frac{6}{7},(\frac{\sqrt{13}}{7})[/tex] and [tex](\frac{5}{13} ,\frac{12}{13})[/tex] are the only two points that  lies on unit circle.

Step-by-step explanation:

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

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

[tex](x)^2+(y)^2=1[/tex]  .........(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) [tex](\frac{1}{3} ,\frac{2}{3})[/tex]

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

[tex](x)^2+(y)^2=1[/tex]

[tex]\Rightarrow (\frac{1}{3})^2+(\frac{2}{3})^2[/tex]

[tex]\Rightarrow (\frac{1}{9})+(\frac{4}{9})[/tex]

[tex]\Rightarrow (\frac{1+4}{9})=\frac{5}{9} \neq 1[/tex]

Thus, [tex](\frac{1}{3} ,\frac{2}{3})[/tex] point do not lies on unit circle.

2) [tex](\frac{4}{3} ,\frac{4}{5})[/tex]

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

[tex](x)^2+(y)^2=1[/tex]

[tex]\Rightarrow (\frac{4}{3})^2+(\frac{4}{5})^2[/tex]

[tex]\Rightarrow (\frac{16}{9})+(\frac{16}{25})[/tex]

[tex]\Rightarrow (\frac{544}{225}) \neq 1[/tex]

Thus,  [tex](\frac{4}{3} ,\frac{4}{5})[/tex] point do not lies on unit circle.

3) [tex](\frac{6}{7} ,\frac{\sqrt{13}}{7})[/tex]

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

[tex](x)^2+(y)^2=1[/tex]

[tex]\Rightarrow (\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2[/tex]

[tex]\Rightarrow (\frac{36}{49})+(\frac{13}{49})[/tex]

[tex]\Rightarrow (\frac{36+13}{49})=\frac{49}{49}=1[/tex]

Thus,[tex](\frac{6}{7} ,\frac{\sqrt{13}}{7})[/tex] point lies on unit circle.

4) [tex](\frac{5}{13} ,\frac{12}{13})[/tex]

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

[tex](x)^2+(y)^2=1[/tex]

[tex]\Rightarrow (\frac{5}{13})^2+(\frac{12}{13})^2[/tex]

[tex]\Rightarrow (\frac{25}{169})+(\frac{144}{169})[/tex]

[tex]\Rightarrow (\frac{25+144}{169})=\frac{169}{169}=1[/tex]

Thus, [tex](\frac{5}{13} ,\frac{12}{13})[/tex] point lies on unit circle.

Thus, only two points lies on unit circle.

[tex](\frac{6}{7} ,\frac{\sqrt{13}}{7})[/tex] and [tex](\frac{5}{13} ,\frac{12}{13})[/tex] are the only two points that  lies on unit circle.


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