Answer :
To show that a point [tex]\((x, y)\)[/tex] is on the unit circle, we need to verify that it satisfies the equation of the unit circle, [tex]\(x^2 + y^2 = 1\)[/tex].
### Point 1: [tex]\(\left(\frac{4}{5}, -\frac{3}{5}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(\frac{4}{5}\right)^2 + \left(-\frac{3}{5}\right)^2 = \frac{4^2}{5^2} + \frac{(-3)^2}{5^2} = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(\frac{4}{5}, -\frac{3}{5}\right)\)[/tex] is on the unit circle.
### Point 2: [tex]\(\left(-\frac{5}{13}, \frac{12}{13}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(-\frac{5}{13}\right)^2 + \left(\frac{12}{13}\right)^2 = \frac{(-5)^2}{13^2} + \frac{12^2}{13^2} = \frac{25}{169} + \frac{144}{169} = \frac{169}{169} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(-\frac{5}{13}, \frac{12}{13}\right)\)[/tex] is on the unit circle.
### Point 3: [tex]\(\left(\frac{7}{25}, \frac{24}{25}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(\frac{7}{25}\right)^2 + \left(\frac{24}{25}\right)^2 = \frac{7^2}{25^2} + \frac{24^2}{25^2} = \frac{49}{625} + \frac{576}{625} = \frac{625}{625} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(\frac{7}{25}, \frac{24}{25}\right)\)[/tex] is on the unit circle.
### Point 4: [tex]\(\left(-\frac{5}{7}, -\frac{2 \sqrt{6}}{7}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(-\frac{5}{7}\right)^2 + \left(-\frac{2 \sqrt{6}}{7}\right)^2 = \frac{(-5)^2}{7^2} + \frac{(2 \sqrt{6})^2}{7^2} = \frac{25}{49} + \frac{24}{49} = \frac{49}{49} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(-\frac{5}{7}, -\frac{2 \sqrt{6}}{7}\right)\)[/tex] is on the unit circle.
### Point 5: [tex]\(\left(-\frac{\sqrt{5}}{3}, \frac{2}{3}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(-\frac{\sqrt{5}}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{(\sqrt{5})^2}{3^2} + \frac{2^2}{3^2} = \frac{5}{9} + \frac{4}{9} = \frac{9}{9} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(-\frac{\sqrt{5}}{3}, \frac{2}{3}\right)\)[/tex] is on the unit circle.
### Point 6: [tex]\(\left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(\frac{\sqrt{11}}{6}\right)^2 + \left(\frac{5}{6}\right)^2 = \frac{(\sqrt{11})^2}{6^2} + \frac{5^2}{6^2} = \frac{11}{36} + \frac{25}{36} = \frac{36}{36} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right)\)[/tex] is on the unit circle.
Therefore, all given points are on the unit circle.
### Point 1: [tex]\(\left(\frac{4}{5}, -\frac{3}{5}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(\frac{4}{5}\right)^2 + \left(-\frac{3}{5}\right)^2 = \frac{4^2}{5^2} + \frac{(-3)^2}{5^2} = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(\frac{4}{5}, -\frac{3}{5}\right)\)[/tex] is on the unit circle.
### Point 2: [tex]\(\left(-\frac{5}{13}, \frac{12}{13}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(-\frac{5}{13}\right)^2 + \left(\frac{12}{13}\right)^2 = \frac{(-5)^2}{13^2} + \frac{12^2}{13^2} = \frac{25}{169} + \frac{144}{169} = \frac{169}{169} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(-\frac{5}{13}, \frac{12}{13}\right)\)[/tex] is on the unit circle.
### Point 3: [tex]\(\left(\frac{7}{25}, \frac{24}{25}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(\frac{7}{25}\right)^2 + \left(\frac{24}{25}\right)^2 = \frac{7^2}{25^2} + \frac{24^2}{25^2} = \frac{49}{625} + \frac{576}{625} = \frac{625}{625} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(\frac{7}{25}, \frac{24}{25}\right)\)[/tex] is on the unit circle.
### Point 4: [tex]\(\left(-\frac{5}{7}, -\frac{2 \sqrt{6}}{7}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(-\frac{5}{7}\right)^2 + \left(-\frac{2 \sqrt{6}}{7}\right)^2 = \frac{(-5)^2}{7^2} + \frac{(2 \sqrt{6})^2}{7^2} = \frac{25}{49} + \frac{24}{49} = \frac{49}{49} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(-\frac{5}{7}, -\frac{2 \sqrt{6}}{7}\right)\)[/tex] is on the unit circle.
### Point 5: [tex]\(\left(-\frac{\sqrt{5}}{3}, \frac{2}{3}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(-\frac{\sqrt{5}}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{(\sqrt{5})^2}{3^2} + \frac{2^2}{3^2} = \frac{5}{9} + \frac{4}{9} = \frac{9}{9} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(-\frac{\sqrt{5}}{3}, \frac{2}{3}\right)\)[/tex] is on the unit circle.
### Point 6: [tex]\(\left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right)\)[/tex]
1. Calculate [tex]\(x^2 + y^2\)[/tex]:
[tex]\[ \left(\frac{\sqrt{11}}{6}\right)^2 + \left(\frac{5}{6}\right)^2 = \frac{(\sqrt{11})^2}{6^2} + \frac{5^2}{6^2} = \frac{11}{36} + \frac{25}{36} = \frac{36}{36} = 1 \][/tex]
2. Since [tex]\(x^2 + y^2 = 1\)[/tex], the point [tex]\(\left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right)\)[/tex] is on the unit circle.
Therefore, all given points are on the unit circle.
