Given the figure, we can deduce the following information:
1. The two given points are:
(1,0)
(x,-7/10)
To determine the value of x, we must note first that the x value on the first quadrant is 1 so the radius of the circle must be equal to 1.
Next, we apply the equation for unit circle:
[tex]x^2+y^2=1[/tex]
Then, we plug in y=-7/10 into x^2+y^2=1:
[tex]\begin{gathered} x^2+y^2=1 \\ x^2+(-\frac{7}{10})^2=1 \\ x^2+\frac{49}{100}=1 \\ \text{Simplify and rearrange} \\ x^2=1-\frac{49}{100} \\ x^2=\frac{51}{100} \\ x=\pm\sqrt[]{\frac{51}{100}} \\ x=\pm\frac{\sqrt[]{51}}{10} \end{gathered}[/tex]
But since the point is in the third quadrant, the value of x must be negative. Therefore, the answer is:
[tex]x=-\frac{\sqrt[]{51}}{10}[/tex]
