Certainly! Let's determine which points could be on the unit circle. A point \((x,y)\) lies on the unit circle if and only if \(x^2 + y^2 = 1\).
1. Point A: \((0.8, -0.6)\)
[tex]\[
x = 0.8, \quad y = -0.6
\][/tex]
Calculate \(x^2 + y^2\):
[tex]\[
x^2 = 0.8^2 = 0.64
\][/tex]
[tex]\[
y^2 = (-0.6)^2 = 0.36
\][/tex]
[tex]\[
x^2 + y^2 = 0.64 + 0.36 = 1.00
\][/tex]
Since \(x^2 + y^2 = 1\), the point \((0.8, -0.6)\) is on the unit circle.
2. Point B: \(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)
[tex]\[
x = \frac{\sqrt{3}}{2}, \quad y = \frac{1}{3}
\][/tex]
Calculate \(x^2 + y^2\):
[tex]\[
x^2 = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}
\][/tex]
[tex]\[
y^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}
\][/tex]
[tex]\[
x^2 + y^2 = \frac{3}{4} + \frac{1}{9}
\][/tex]
To add these fractions, find a common denominator:
[tex]\[
\frac{3}{4} = \frac{3 \times 9}{4 \times 9} = \frac{27}{36}
\][/tex]
[tex]\[
\frac{1}{9} = \frac{1 \times 4}{9 \times 4} = \frac{4}{36}
\][/tex]
[tex]\[
x^2 + y^2 = \frac{27}{36} + \frac{4}{36} = \frac{31}{36}
\][/tex]
Since \(x^2 + y^2 = \frac{31}{36}\), which is not equal to 1, the point \(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\) is not on the unit circle.
3. Point C: \((1, 1)\)
[tex]\[
x = 1, \quad y = 1
\][/tex]
Calculate \(x^2 + y^2\):
[tex]\[
x^2 = 1^2 = 1
\][/tex]
[tex]\[
y^2 = 1^2 = 1
\][/tex]
[tex]\[
x^2 + y^2 = 1 + 1 = 2
\][/tex]
Since \(x^2 + y^2 = 2\), the point \((1, 1)\) is not on the unit circle.
4. Point D: \(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)
[tex]\[
x = -\frac{2}{3}, \quad y = \frac{\sqrt{5}}{3}
\][/tex]
Calculate \(x^2 + y^2\):
[tex]\[
x^2 = \left(-\frac{2}{3}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}
\][/tex]
[tex]\[
y^2 = \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{5}{9}
\][/tex]
[tex]\[
x^2 + y^2 = \frac{4}{9} + \frac{5}{9} = \frac{9}{9} = 1
\][/tex]
Since \(x^2 + y^2 = 1\), the point \(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\) is on the unit circle.
Hence, the points that could be on the unit circle are:
- A. \((0.8, -0.6)\)
- D. \(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)
These are the points that lie on the unit circle.
