Respuesta :
Answer:
1. [tex]y=\frac{-3}{5}[/tex]
2. [tex]x=\frac{-\sqrt{5}}{3}[/tex]
Step-by-step explanation:
Ques 1: We are given that the point [tex]P(\frac{-4}{5},y)[/tex] lies on a unit circle and is in the 3rd quadrant.
The equation of the unit circle is [tex]x^{2}+y^{2}=1[/tex].
Substituting the values, we get,
[tex]x^{2}+y^{2}=1[/tex]
⇒ [tex](\frac{-4}{5})^{2}+y^{2}=1[/tex]
⇒ [tex]y^{2}=1-(\frac{-4}{5})^{2}[/tex]
⇒ [tex]y^{2}=1-\frac{16}{25}[/tex]
⇒ [tex]y^{2}=\frac{25-16}{25}[/tex]
⇒ [tex]y^{2}=\frac{9}{25}[/tex]
⇒ [tex]y=\pm \frac{3}{5}[/tex]
Since, the point P lies in the 3rd quadrant i.e. the value of y will be negative.
So, [tex]y=\frac{-3}{5}[/tex].
Ques 2: We are given that the point [tex]P(x,\frac{2}{3})[/tex] lies on a unit circle and is in the 2nd quadrant.
The equation of the unit circle is [tex]x^{2}+y^{2}=1[/tex].
Substituting the values, we get,
[tex]x^{2}+y^{2}=1[/tex]
⇒ [tex]x^{2}+(\frac{2}{3})^{2}=1[/tex]
⇒ [tex]x^{2}=1-(\frac{2}{3})^{2}[/tex]
⇒ [tex]x^{2}=1-\frac{4}{9}[/tex]
⇒ [tex]x^{2}=\frac{9-4}{9}[/tex]
⇒ [tex]x^{2}=\frac{5}{9}[/tex]
⇒ [tex]x=\pm \frac{\sqrt{5}}{3}[/tex]
Since, the point P lies in the 2nd quadrant i.e. the value of x will be negative.
So, [tex]x=\frac{-\sqrt{5}}{3}[/tex]
