Math Problem Statement

Is each point on the unit circle? How do you know?

Solution

The problem is asking whether each point provided in the question is on the unit circle. A point (x,y)(x, y) lies on the unit circle if and only if it satisfies the equation:

x2+y2=1x^2 + y^2 = 1

Let’s evaluate each of the points given in the problem:

a) (34,14)\left( -\frac{3}{4}, \frac{1}{4} \right)

x2+y2=(34)2+(14)2=916+116=1016=58x^2 + y^2 = \left( -\frac{3}{4} \right)^2 + \left( \frac{1}{4} \right)^2 = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8} 581(Not on the unit circle)\frac{5}{8} \neq 1 \quad (\text{Not on the unit circle})

b) (58,78)\left( \frac{\sqrt{5}}{8}, \frac{7}{8} \right)

x2+y2=(58)2+(78)2=564+4964=5464=2732x^2 + y^2 = \left( \frac{\sqrt{5}}{8} \right)^2 + \left( \frac{7}{8} \right)^2 = \frac{5}{64} + \frac{49}{64} = \frac{54}{64} = \frac{27}{32} 27321(Not on the unit circle)\frac{27}{32} \neq 1 \quad (\text{Not on the unit circle})

c) (513,1213)\left( -\frac{5}{13}, \frac{12}{13} \right)

x2+y2=(513)2+(1213)2=25169+144169=169169=1x^2 + y^2 = \left( -\frac{5}{13} \right)^2 + \left( \frac{12}{13} \right)^2 = \frac{25}{169} + \frac{144}{169} = \frac{169}{169} = 1 1=1(On the unit circle)1 = 1 \quad (\text{On the unit circle})

d) (45,35)\left( \frac{4}{5}, -\frac{3}{5} \right)

x2+y2=(45)2+(35)2=1625+925=2525=1x^2 + y^2 = \left( \frac{4}{5} \right)^2 + \left( -\frac{3}{5} \right)^2 = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1 1=1(On the unit circle)1 = 1 \quad (\text{On the unit circle})

e) (32,12)\left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right)

x2+y2=(32)2+(12)2=34+14=1x^2 + y^2 = \left( -\frac{\sqrt{3}}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 = \frac{3}{4} + \frac{1}{4} = 1 1=1(On the unit circle)1 = 1 \quad (\text{On the unit circle})

f) (74,34)\left( \frac{\sqrt{7}}{4}, \frac{3}{4} \right)

x2+y2=(74)2+(34)2=716+916=1616=1x^2 + y^2 = \left( \frac{\sqrt{7}}{4} \right)^2 + \left( \frac{3}{4} \right)^2 = \frac{7}{16} + \frac{9}{16} = \frac{16}{16} = 1 1=1(On the unit circle)1 = 1 \quad (\text{On the unit circle})

Conclusion:

  • a) Not on the unit circle
  • b) Not on the unit circle
  • c) On the unit circle
  • d) On the unit circle
  • e) On the unit circle
  • f) On the unit circle

Do you want more details or have any questions about this?

Related Questions:

  1. How can you derive the equation of the unit circle?
  2. How can you use trigonometric identities to find points on the unit circle?
  3. What is the relationship between a point on the unit circle and the corresponding angle in radians?
  4. How do parametric equations describe points on the unit circle?
  5. What are common points on the unit circle and their coordinates?

Tip:

Whenever checking if a point lies on the unit circle, always apply the equation x2+y2=1x^2 + y^2 = 1.

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Math Problem Analysis

Mathematical Concepts

Geometry
Trigonometry

Formulas

Unit circle equation: x^2 + y^2 = 1

Theorems

Unit Circle Theorem

Suitable Grade Level

Grades 10-12