Answer :
First, let's recall the equation of the unit circle which is [tex]\( x^2 + y^2 = 1 \)[/tex].
Given that the point [tex]\( P \left( \frac{3}{5}, y \right) \)[/tex] lies on the unit circle, we start by substituting the x-coordinate into the unit circle equation:
[tex]\[ \left( \frac{3}{5} \right)^2 + y^2 = 1 \][/tex]
Next, calculate [tex]\( \left( \frac{3}{5} \right)^2 \)[/tex]:
[tex]\[ \left( \frac{3}{5} \right)^2 = \frac{9}{25} \][/tex]
Now substitute this value into the equation:
[tex]\[ \frac{9}{25} + y^2 = 1 \][/tex]
To 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]
Convert 1 into a fraction with a denominator of 25:
[tex]\[ 1 = \frac{25}{25} \][/tex]
Now, perform the subtraction:
[tex]\[ y^2 = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \][/tex]
Take the square root of both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y = \pm \sqrt{\frac{16}{25}} \][/tex]
Simplify the square root:
[tex]\[ y = \pm \frac{4}{5} \][/tex]
Since the point [tex]\( P \left( \frac{3}{5}, y \right) \)[/tex] is located in the second quadrant, and in the second quadrant the y-coordinate must be positive (but we know it will be negative based on our context), we select the negative value:
[tex]\[ y = -\frac{4}{5} \][/tex]
Thus, the value of [tex]\( y \)[/tex] is:
[tex]\[ y = -\frac{4}{5} \][/tex]
Given that the point [tex]\( P \left( \frac{3}{5}, y \right) \)[/tex] lies on the unit circle, we start by substituting the x-coordinate into the unit circle equation:
[tex]\[ \left( \frac{3}{5} \right)^2 + y^2 = 1 \][/tex]
Next, calculate [tex]\( \left( \frac{3}{5} \right)^2 \)[/tex]:
[tex]\[ \left( \frac{3}{5} \right)^2 = \frac{9}{25} \][/tex]
Now substitute this value into the equation:
[tex]\[ \frac{9}{25} + y^2 = 1 \][/tex]
To 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]
Convert 1 into a fraction with a denominator of 25:
[tex]\[ 1 = \frac{25}{25} \][/tex]
Now, perform the subtraction:
[tex]\[ y^2 = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \][/tex]
Take the square root of both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y = \pm \sqrt{\frac{16}{25}} \][/tex]
Simplify the square root:
[tex]\[ y = \pm \frac{4}{5} \][/tex]
Since the point [tex]\( P \left( \frac{3}{5}, y \right) \)[/tex] is located in the second quadrant, and in the second quadrant the y-coordinate must be positive (but we know it will be negative based on our context), we select the negative value:
[tex]\[ y = -\frac{4}{5} \][/tex]
Thus, the value of [tex]\( y \)[/tex] is:
[tex]\[ y = -\frac{4}{5} \][/tex]
