Answer: [tex]\bold{x=\frac{2 \sqrt{10}}{7}}[/tex]
Step-by-step explanation:
[tex]\text { Let } p=\left(x,-\frac{3}{7}\right) \text { as given } y \text {-coordinate is }=-\frac{3}{7}[/tex]
[tex]r=1 \ \ \text{(as $p$ is on unit circle)} \\$\therefore x^{2}+y^{2}=1^{2}[/tex]
[tex]\begin{aligned}&\Rightarrow x^{2}+\left(-\frac{3}{7}\right)^{2}=1 \Rightarrow x^{2}=1-\frac{9}{49}=\frac{40}{49} \\&\Rightarrow x^{2}=\pm \sqrt{\frac{40}{49}} \Rightarrow x=\pm \frac{2 \sqrt{10}}{7}\end{aligned}[/tex]
[tex]$Since the point is in quadrant $4$ $\therefore \text{the} \ x$-coordinate will be positive$\therefore x=\frac{2 \sqrt{10}}{7} \text$[/tex]
