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Sagot :
To determine which points lie on the unit circle, we need to verify if each point [tex]\((x, y)\)[/tex] satisfies the equation of the unit circle, [tex]\( x^2 + y^2 = 1 \)[/tex].
Let's check each point step-by-step:
Point A: [tex]\((1, 1)\)[/tex]
1. Compute [tex]\( x^2 \)[/tex] and [tex]\( y^2 \)[/tex]:
[tex]\[ x^2 = 1^2 = 1 \][/tex]
[tex]\[ y^2 = 1^2 = 1 \][/tex]
2. Add the squares:
[tex]\[ x^2 + y^2 = 1 + 1 = 2 \][/tex]
Since [tex]\( 2 \neq 1 \)[/tex], the point [tex]\((1, 1)\)[/tex] does not lie on the unit circle.
Point B: [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex]
1. Compute [tex]\( x^2 \)[/tex] and [tex]\( y^2 \)[/tex]:
[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]
2. Add the squares:
[tex]\[ x^2 + y^2 = \frac{3}{4} + \frac{1}{9} \][/tex]
To add these, we need a common denominator, which is 36:
[tex]\[ x^2 + y^2 = \frac{27}{36} + \frac{4}{36} = \frac{31}{36} \][/tex]
Since [tex]\( \frac{31}{36} \neq 1 \)[/tex], the point [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex] does not lie on the unit circle.
Point C: [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex]
1. Compute [tex]\( x^2 \)[/tex] and [tex]\( y^2 \)[/tex]:
[tex]\[ x^2 = \left( -\frac{2}{3} \right)^2 = \frac{4}{9} \][/tex]
[tex]\[ y^2 = \left( \frac{\sqrt{5}}{3} \right)^2 = \frac{5}{9} \][/tex]
2. Add the squares:
[tex]\[ x^2 + y^2 = \frac{4}{9} + \frac{5}{9} = \frac{9}{9} = 1 \][/tex]
Since [tex]\( 1 = 1 \)[/tex], the point [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex] lies on the unit circle.
Point D: [tex]\((0.8, -0.6)\)[/tex]
1. Compute [tex]\( x^2 \)[/tex] and [tex]\( y^2 \)[/tex]:
[tex]\[ x^2 = 0.8^2 = 0.64 \][/tex]
[tex]\[ y^2 = (-0.6)^2 = 0.36 \][/tex]
2. Add the squares:
[tex]\[ x^2 + y^2 = 0.64 + 0.36 = 1 \][/tex]
Since [tex]\( 1 = 1 \)[/tex], the point [tex]\((0.8, -0.6)\)[/tex] lies on the unit circle.
Answer:
The points that lie on the unit circle are:
C. [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex]
D. [tex]\((0.8, -0.6)\)[/tex]
Let's check each point step-by-step:
Point A: [tex]\((1, 1)\)[/tex]
1. Compute [tex]\( x^2 \)[/tex] and [tex]\( y^2 \)[/tex]:
[tex]\[ x^2 = 1^2 = 1 \][/tex]
[tex]\[ y^2 = 1^2 = 1 \][/tex]
2. Add the squares:
[tex]\[ x^2 + y^2 = 1 + 1 = 2 \][/tex]
Since [tex]\( 2 \neq 1 \)[/tex], the point [tex]\((1, 1)\)[/tex] does not lie on the unit circle.
Point B: [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex]
1. Compute [tex]\( x^2 \)[/tex] and [tex]\( y^2 \)[/tex]:
[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]
2. Add the squares:
[tex]\[ x^2 + y^2 = \frac{3}{4} + \frac{1}{9} \][/tex]
To add these, we need a common denominator, which is 36:
[tex]\[ x^2 + y^2 = \frac{27}{36} + \frac{4}{36} = \frac{31}{36} \][/tex]
Since [tex]\( \frac{31}{36} \neq 1 \)[/tex], the point [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{3}\right)\)[/tex] does not lie on the unit circle.
Point C: [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex]
1. Compute [tex]\( x^2 \)[/tex] and [tex]\( y^2 \)[/tex]:
[tex]\[ x^2 = \left( -\frac{2}{3} \right)^2 = \frac{4}{9} \][/tex]
[tex]\[ y^2 = \left( \frac{\sqrt{5}}{3} \right)^2 = \frac{5}{9} \][/tex]
2. Add the squares:
[tex]\[ x^2 + y^2 = \frac{4}{9} + \frac{5}{9} = \frac{9}{9} = 1 \][/tex]
Since [tex]\( 1 = 1 \)[/tex], the point [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex] lies on the unit circle.
Point D: [tex]\((0.8, -0.6)\)[/tex]
1. Compute [tex]\( x^2 \)[/tex] and [tex]\( y^2 \)[/tex]:
[tex]\[ x^2 = 0.8^2 = 0.64 \][/tex]
[tex]\[ y^2 = (-0.6)^2 = 0.36 \][/tex]
2. Add the squares:
[tex]\[ x^2 + y^2 = 0.64 + 0.36 = 1 \][/tex]
Since [tex]\( 1 = 1 \)[/tex], the point [tex]\((0.8, -0.6)\)[/tex] lies on the unit circle.
Answer:
The points that lie on the unit circle are:
C. [tex]\(\left(-\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\)[/tex]
D. [tex]\((0.8, -0.6)\)[/tex]
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