Question

if point p(-12/3,y) lies on the unit circle and P is in the second quadrant y is

Answer 1

not possible for that point to lie on the unit circle because \((\frac{-12}{3})^2>1\)

Answer 2

i am going to make a guess that it is \((-\frac{12}{13},y)\)

Answer 3

yes i meant 13

Answer 4

then find the third side by pythagoras

Answer 5

\[12^2+y^2=13^2\]
\[y=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\]
of course we are on the unit circle, so you have to divide by 13 and therefore \(y=\frac{5}{13}\)

Answer 6

and it is positive not negative, because in the second quadrant you are above the x - axis