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To determine which points could not lie on the unit circle, we need to check if they satisfy the equation of the unit circle. The equation for a unit circle centered at the origin is given by:
[tex]\[ x^2 + y^2 = 1 \][/tex]
We will substitute each pair of [tex]\((x, y)\)[/tex] into this equation and check if the left-hand side equals 1.
### Point A: [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex]
Substitute [tex]\(x = -\frac{2}{3}\)[/tex] and [tex]\(y = \frac{\sqrt{5}}{3}\)[/tex]:
[tex]\[ \left(-\frac{2}{3}\right)^2 + \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{4}{9} + \frac{5}{9} = \frac{4+5}{9} = \frac{9}{9} = 1 \][/tex]
Since [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex] satisfies the unit circle equation, this point lies on the unit circle.
### Point B: [tex]\((1,1)\)[/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 1\)[/tex]:
[tex]\[ 1^2 + 1^2 = 1 + 1 = 2 \][/tex]
Since [tex]\( 2 \neq 1 \)[/tex], the point [tex]\((1, 1)\)[/tex] does not satisfy the unit circle equation. Therefore, this point does not lie on the unit circle.
### Point C: [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex]
Substitute [tex]\(x = \frac{\sqrt{3}}{2}\)[/tex] and [tex]\(y = \frac{1}{3}\)[/tex]:
[tex]\[ \left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{3} \right)^2 = \frac{3}{4} + \frac{1}{9} \][/tex]
Convert to a common denominator:
[tex]\[ \frac{3}{4} = \frac{27}{36} \][/tex]
[tex]\[ \frac{1}{9} = \frac{4}{36} \][/tex]
[tex]\[ \frac{27}{36} + \frac{4}{36} = \frac{31}{36} \][/tex]
Since [tex]\( \frac{31}{36} \neq 1 \)[/tex], the point [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex] does not satisfy the unit circle equation. Therefore, this point does not lie on the unit circle.
### Point D: [tex]\((0.8, -0.6)\)[/tex]
Substitute [tex]\(x = 0.8\)[/tex] and [tex]\(y = -0.6\)[/tex]:
[tex]\[ (0.8)^2 + (-0.6)^2 = 0.64 + 0.36 = 1 \][/tex]
Since [tex]\((0.8, -0.6)\)[/tex] satisfies the unit circle equation, this point lies on the unit circle.
### Conclusion
The points provided in the options are:
- A. [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex]
- B. [tex]\((1, 1)\)[/tex]
- C. [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex]
- D. [tex]\((0.8, -0.6)\)[/tex]
From our calculations, the points that do not lie on the unit circle are:
- B. [tex]\((1, 1)\)[/tex]
- C. [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex]
Therefore, the points B and C could not be points on the unit circle.
[tex]\[ x^2 + y^2 = 1 \][/tex]
We will substitute each pair of [tex]\((x, y)\)[/tex] into this equation and check if the left-hand side equals 1.
### Point A: [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex]
Substitute [tex]\(x = -\frac{2}{3}\)[/tex] and [tex]\(y = \frac{\sqrt{5}}{3}\)[/tex]:
[tex]\[ \left(-\frac{2}{3}\right)^2 + \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{4}{9} + \frac{5}{9} = \frac{4+5}{9} = \frac{9}{9} = 1 \][/tex]
Since [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex] satisfies the unit circle equation, this point lies on the unit circle.
### Point B: [tex]\((1,1)\)[/tex]
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 1\)[/tex]:
[tex]\[ 1^2 + 1^2 = 1 + 1 = 2 \][/tex]
Since [tex]\( 2 \neq 1 \)[/tex], the point [tex]\((1, 1)\)[/tex] does not satisfy the unit circle equation. Therefore, this point does not lie on the unit circle.
### Point C: [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex]
Substitute [tex]\(x = \frac{\sqrt{3}}{2}\)[/tex] and [tex]\(y = \frac{1}{3}\)[/tex]:
[tex]\[ \left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{3} \right)^2 = \frac{3}{4} + \frac{1}{9} \][/tex]
Convert to a common denominator:
[tex]\[ \frac{3}{4} = \frac{27}{36} \][/tex]
[tex]\[ \frac{1}{9} = \frac{4}{36} \][/tex]
[tex]\[ \frac{27}{36} + \frac{4}{36} = \frac{31}{36} \][/tex]
Since [tex]\( \frac{31}{36} \neq 1 \)[/tex], the point [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex] does not satisfy the unit circle equation. Therefore, this point does not lie on the unit circle.
### Point D: [tex]\((0.8, -0.6)\)[/tex]
Substitute [tex]\(x = 0.8\)[/tex] and [tex]\(y = -0.6\)[/tex]:
[tex]\[ (0.8)^2 + (-0.6)^2 = 0.64 + 0.36 = 1 \][/tex]
Since [tex]\((0.8, -0.6)\)[/tex] satisfies the unit circle equation, this point lies on the unit circle.
### Conclusion
The points provided in the options are:
- A. [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex]
- B. [tex]\((1, 1)\)[/tex]
- C. [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex]
- D. [tex]\((0.8, -0.6)\)[/tex]
From our calculations, the points that do not lie on the unit circle are:
- B. [tex]\((1, 1)\)[/tex]
- C. [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex]
Therefore, the points B and C could not be points on the unit circle.
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