If the point [tex]P\left(-\frac{3}{5}, y\right)[/tex] lies on the unit circle and [tex]P[/tex] is in the second quadrant, what does [tex]\( y \)[/tex] equal?

If necessary, use the slash mark (/) for a fraction bar.



Answer :

To determine the value of [tex]\( y \)[/tex] for the point [tex]\( P\left(-\frac{3}{5}, y\right) \)[/tex] that lies on the unit circle in the second quadrant, we start with the equation of a unit circle:

[tex]\[ x^2 + y^2 = 1 \][/tex]

Given [tex]\( x = -\frac{3}{5} \)[/tex], substitute this value into the equation:

[tex]\[ \left(-\frac{3}{5}\right)^2 + y^2 = 1 \][/tex]

Now, calculate [tex]\( \left(-\frac{3}{5}\right)^2 \)[/tex]:

[tex]\[ \left(-\frac{3}{5}\right)^2 = \frac{9}{25} \][/tex]

Next, substitute [tex]\( \frac{9}{25} \)[/tex] in place of [tex]\( \left(-\frac{3}{5}\right)^2 \)[/tex]:

[tex]\[ \frac{9}{25} + y^2 = 1 \][/tex]

To solve for [tex]\( y^2 \)[/tex], subtract [tex]\( \frac{9}{25} \)[/tex] from both sides of the equation:

[tex]\[ y^2 = 1 - \frac{9}{25} \][/tex]

Convert 1 into a fraction with the same denominator:

[tex]\[ 1 = \frac{25}{25} \][/tex]

Now perform the subtraction:

[tex]\[ y^2 = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \][/tex]

Thus,

[tex]\[ y^2 = \frac{16}{25} \][/tex]

To find [tex]\( y \)[/tex], take the square root of both sides:

[tex]\[ y = \sqrt{\frac{16}{25}} \][/tex]

Since [tex]\( y \)[/tex] is in the second quadrant, where [tex]\( y \)[/tex] values are positive:

[tex]\[ y = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} \][/tex]

Therefore, the value of [tex]\( y \)[/tex] is:

[tex]\[ y = \frac{4}{5} \][/tex]

Thus, [tex]\( y = 0.8 \)[/tex]