Given that the point [tex]\( P\left(-\frac{3}{5}, y\right) \)[/tex] lies on the unit circle, we know it must satisfy the equation of the unit circle:
[tex]\[ x^2 + y^2 = 1 \][/tex]
Let's break down the steps to find [tex]\( y \)[/tex]:
1. Set up the given equation:
Given [tex]\( x = -\frac{3}{5} \)[/tex], substitute [tex]\( x \)[/tex] into the unit circle equation:
[tex]\[ \left(-\frac{3}{5}\right)^2 + y^2 = 1 \][/tex]
2. Simplify the equation:
Calculate [tex]\( \left(-\frac{3}{5}\right)^2 \)[/tex]:
[tex]\[ \left(-\frac{3}{5}\right)^2 = \frac{9}{25} \][/tex]
So the equation becomes:
[tex]\[ \frac{9}{25} + y^2 = 1 \][/tex]
3. Isolate [tex]\( y^2 \)[/tex]:
Subtract [tex]\(\frac{9}{25}\)[/tex] from both sides of the equation:
[tex]\[ y^2 = 1 - \frac{9}{25} \][/tex]
4. Simplify the right-hand side:
Rewrite [tex]\( 1 \)[/tex] as [tex]\(\frac{25}{25}\)[/tex] to have common denominators:
[tex]\[ y^2 = \frac{25}{25} - \frac{9}{25} \][/tex]
Subtract the fractions:
[tex]\[ y^2 = \frac{25 - 9}{25} = \frac{16}{25} \][/tex]
5. Solve for [tex]\( y \)[/tex]:
Take the square root of both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y = \pm\sqrt{\frac{16}{25}} \][/tex]
[tex]\[ y = \pm\frac{\sqrt{16}}{\sqrt{25}} \][/tex]
[tex]\[ y = \pm\frac{4}{5} \][/tex]
6. Determine the correct sign:
Since point [tex]\( P \)[/tex] is in the second quadrant, and in that quadrant [tex]\( x \)[/tex] is negative (as given) and [tex]\( y \)[/tex] is positive, we select the positive value of [tex]\( y \)[/tex]:
[tex]\[ y = \frac{4}{5} \][/tex]
Therefore, the value of [tex]\( y \)[/tex] is:
[tex]\[ y = \frac{4}{5} \][/tex]
