Respuesta :
Answer:
Option (2) and (4) are correct.
[tex](\frac{5}{13},\frac{12}{13} )[/tex] and [tex](\frac{6}{7},\frac{\sqrt{13}}{7} )[/tex] are points on the unit circle.
Step-by-step explanation:
Given : Some points of circles.
We have to choose which points could be points on the unit circle.
We know, the equation of circle is
[tex]x^2+y^2=r^2[/tex] ..........(1)
where r is radius of circle.
We check each point for x and y values on the equation of circle and see which point gives radius = 1
Thus,
1) [tex](\frac{4}{3},\frac{4}{5} )[/tex]
Put in LHS of (1) , we have,
[tex](\frac{4}{3})^2+(\frac{4}{5})^2[/tex]
Simplify, we have,
[tex]=\frac{16}{9}+\frac{16}{25}[/tex]
[tex]=\frac{544}{225}\neq 1[/tex]
Thus, [tex](\frac{4}{3},\frac{4}{5} )[/tex] is not a point on the unit circle.
2) [tex](\frac{5}{13},\frac{12}{13} )[/tex]
Put in LHS of (1) , we have,
[tex](\frac{5}{13})^2+(\frac{12}{13})^2[/tex]
Simplify, we have,
[tex]=\frac{25}{169}+\frac{144}{169}[/tex]
[tex]=\frac{169}{169}= 1[/tex]
Thus, [tex](\frac{5}{13},\frac{12}{13} )[/tex] is a point on the unit circle.
3) [tex](\frac{1}{3},\frac{2}{3} )[/tex]
Put in LHS of (1) , we have,
[tex](\frac{1}{3})^2+(\frac{2}{3})^2[/tex]
Simplify, we have,
[tex]=\frac{1}{9}+\frac{4}{9}[/tex]
[tex]=\frac{5}{9}\neq 1[/tex]
Thus, [tex](\frac{1}{3},\frac{2}{3} )[/tex] is not a point on the unit circle.
4) [tex](\frac{6}{7},\frac{\sqrt{13}}{7} )[/tex]
Put in LHS of (1) , we have,
[tex](\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2[/tex]
Simplify, we have,
[tex]=\frac{36}{49}+\frac{13}{49}[/tex]
[tex]=\frac{49}{49}= 1[/tex]
Thus, [tex](\frac{6}{7},\frac{\sqrt{13}}{7} )[/tex] is a point on the unit circle.
Thus, [tex](\frac{5}{13},\frac{12}{13} )[/tex] and [tex](\frac{6}{7},\frac{\sqrt{13}}{7} )[/tex] are points on the unit circle.
You can use the fact that a unit circle has length of radius = 1.
Point which can be on a unit circle is given by
- Option B. (5/13, 12/13)
- Option D: 6/7, [tex]\sqrt{13}/7}[/tex]
What is a unit circle?
A unit circle is a circle with unit (1 sized) radius.
Usually, when we talk of unit circle, we mean it to be centered on origin (0,0) point.
How to find if a point is on unit circle?
If a point is on the unit circle, then the length of the line segment which joins origin with that point will be of 1 unit.
Suppose the point be (x,y), then we have:
[tex]length = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}[/tex]
Using the above conclusion to find which point is lying on the unit circle(assuming centered on origin)
Evaluating distance from origin for all given points one by one
- Option A. (4/3, 4/5)
length = [tex]\sqrt{(4/3)^2 + (4/5)^2} = \dfrac{4\sqrt{34}}{15} \neq 1[/tex]
- Option B. (5/13, 12/13)
length = [tex]\sqrt{(5/13)^2 + (12/13)^2} = \sqrt{169/169} = 1[/tex]
Thus, this point lies on the unit circle.
- Option C. (1/3, 2/3)
length = [tex]\sqrt{(1/3)^2 + (2/3)^2 } = \dfrac{\sqrt{5}}{3} \neq 1[/tex]
- Option D: 6/7, [tex]\sqrt{13}/7}[/tex]
length = [tex]\sqrt{(6/7)^2 + (\sqrt{13}/7})^2} = \sqrt{49/49} = 1[/tex]
Thus, this point lies on the unit circle.
Thus, only option B's point and D's point lies on the unit circle out of given points.
Thus,
Point which can be on a unit circle is given by
- Option B. (5/13, 12/13)
- Option D: 6/7, [tex]\sqrt{13}/7}[/tex]
Learn more about circle here:
https://brainly.com/question/13004063
