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)
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:
\((\frac{6}{7},(\frac{\sqrt{13}}{7})\) and \((\frac{5}{13} ,\frac{12}{13})\) are the only two points that lies on unit circle.
Step-by-step 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) \((\frac{1}{3} ,\frac{2}{3})\)
Substitute for x and y in Left side of (1), we get,
\((x)^2+(y)^2=1\)
\(\Rightarrow (\frac{1}{3})^2+(\frac{2}{3})^2\)
\(\Rightarrow (\frac{1}{9})+(\frac{4}{9})\)
\(\Rightarrow (\frac{1+4}{9})=\frac{5}{9} eq 1\)
Thus, \((\frac{1}{3} ,\frac{2}{3})\) point do not lies on unit circle.
2) \((\frac{4}{3} ,\frac{4}{5})\)
Substitute for x and y in Left side of (1), we get,
\((x)^2+(y)^2=1\)
\(\Rightarrow (\frac{4}{3})^2+(\frac{4}{5})^2\)
\(\Rightarrow (\frac{16}{9})+(\frac{16}{25})\)
\(\Rightarrow (\frac{544}{225}) eq 1\)
Thus, \((\frac{4}{3} ,\frac{4}{5})\) point do not lies on unit circle.
3) \((\frac{6}{7} ,\frac{\sqrt{13}}{7})\)
Substitute for x and y in Left side of (1), we get,
\((x)^2+(y)^2=1\)
\(\Rightarrow (\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2\)
\(\Rightarrow (\frac{36}{49})+(\frac{13}{49})\)
\(\Rightarrow (\frac{36+13}{49})=\frac{49}{49}=1\)
Thus,\((\frac{6}{7} ,\frac{\sqrt{13}}{7})\) point lies on unit circle.
4) \((\frac{5}{13} ,\frac{12}{13})\)
Substitute for x and y in Left side of (1), we get,
\((x)^2+(y)^2=1\)
\(\Rightarrow (\frac{5}{13})^2+(\frac{12}{13})^2\)
\(\Rightarrow (\frac{25}{169})+(\frac{144}{169})\)
\(\Rightarrow (\frac{25+144}{169})=\frac{169}{169}=1\)
Thus, \((\frac{5}{13} ,\frac{12}{13})\) point lies on unit circle.
Thus, only two points lies on unit circle.
\((\frac{6}{7} ,\frac{\sqrt{13}}{7})\) and \((\frac{5}{13} ,\frac{12}{13})\) are the only two points that lies on unit circle.
