Since [tex]P[/tex] lies in the third quadrant, [tex]y[/tex] must be negative. Also, because [tex]P[/tex] lies on the unit circle, its coordinates must satsify
[tex]\left(-\dfrac35\right)^2+y^2=1\implies y^2=\dfrac{16}{25}\implies y=\pm\dfrac45[/tex]
Take the negative root, since [tex]y<0[/tex], so that [tex]y=-\dfrac45[/tex].
