Angle $\theta$ lies in quadrant II with point $A(-4, 6)$ on the terminal arm. Point $P$ is the point of intersection of the terminal arm of $\theta$ and the unit circle centered at $(0,0)$. Determine the exact value of the $X$ coordinate of point $P$.
Not sure if I'm attempting it correctly... and not sure what to do next if I'm wrong. Could you explain and show the correct reasoning?
My Attempt: For the terminal arm, since one end is at the origin, the intercept value $b$ of the terminal arm's straight-line equation $$y = mx + b$$ must be $b = 0$. When $y = 6$ for this line, $x = -4$ so that the slope value $m$ must be such that $$6 = m\cdot(-4)$$ Thus $m = -3/2$ and $y = (-3/2)\cdot x$ for all points of the terminal arm's line. For the unit circle, $x^2 + y^2 = 1$. Substitute for $y$ from the straight-line equation into the circle's equation and get $x^2 + [(-3/2)x]^2 = 1$.