Question

1. If the point P(-4/5, y) lies on the unit circle and P is in the third quadrant, what is y? Explain.

2. If the point P(x, 2/3) lies on the unit circle and P is in the second quadrant, what is x? Explain.

Answer 1

1.
`y=(-3)/(5)`

2.
`x=(-√(5))/(3)`

Explanation:

Ques 1: We are given that the point
`P((-4)/(5),y)` lies on a unit circle and is in the 3rd quadrant.

The equation of the unit circle is
`x^(2)+y^(2)=1`.

Substituting the values, we get,

`x^(2)+y^(2)=1`

⇒
`((-4)/(5))^(2)+y^(2)=1`

⇒
`y^(2)=1-((-4)/(5))^(2)`

⇒
`y^(2)=1-(16)/(25)`

⇒
`y^(2)=(25-16)/(25)`

⇒
`y^(2)=(9)/(25)`

⇒
`y=\pm (3)/(5)`

Since, the point P lies in the 3rd quadrant i.e. the value of y will be negative.

So,
`y=(-3)/(5)`.

Ques 2: We are given that the point
`P(x,(2)/(3))` lies on a unit circle and is in the 2nd quadrant.

The equation of the unit circle is
`x^(2)+y^(2)=1`.

Substituting the values, we get,

`x^(2)+y^(2)=1`

⇒
`x^(2)+((2)/(3))^(2)=1`

⇒
`x^(2)=1-((2)/(3))^(2)`

⇒
`x^(2)=1-(4)/(9)`

⇒
`x^(2)=(9-4)/(9)`

⇒
`x^(2)=(5)/(9)`

⇒
`x=\pm (√(5))/(3)`

Since, the point P lies in the 2nd quadrant i.e. the value of x will be negative.

So,
`x=(-√(5))/(3)`