Question

The point P is on the unit circle. If the y-coordinate of P is -3/7

and P is in quadrant IV, then
T=
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Answer 1

Answer:
`\bold{x=(2 √(10))/(7)}`

Explanation:

`\text { Let } p=\left(x,-(3)/(7)\right) \text { as given } y \text {-coordinate is }=-(3)/(7)`

`r=1 \ \ \text{(as $p$ is on unit circle)} \\$$\therefore x^(2)+y^(2)=1^(2)`

`\begin{aligned}&\Rightarrow x^(2)+\left(-(3)/(7)\right)^(2)=1 \Rightarrow x^(2)=1-(9)/(49)=(40)/(49) \\&\Rightarrow x^(2)=\pm \sqrt{(40)/(49)} \Rightarrow x=\pm (2 √(10))/(7)\end{aligned}`

`$$Since the point is in quadrant $4$ $\therefore \text{the} \ x$-coordinate will be positive$$\therefore x=(2 √(10))/(7) \text$$`