Question

show that the point is on the unit circle.
(3/5, - 4/5)
we need to show that the point satisfies the equation of the unit circle, that is, x² + y² =
x² + y² = ( )² + (- 4/5)²
= + 16/25
=
hence, the point ---select--- on the unit circle.
resources

Explanation:

Step1: Recall unit - circle equation

The equation of the unit circle is $x^{2}+y^{2}=1$.

Step2: Identify x and y values

Given the point $(\frac{3}{5},-\frac{4}{5})$, we have $x = \frac{3}{5}$ and $y=-\frac{4}{5}$.

Step3: Substitute x and y into the equation

$x^{2}+y^{2}=(\frac{3}{5})^{2}+(-\frac{4}{5})^{2}$.

Step4: Calculate $(\frac{3}{5})^{2}$

$(\frac{3}{5})^{2}=\frac{3^{2}}{5^{2}}=\frac{9}{25}$.

Step5: Calculate the sum

$x^{2}+y^{2}=\frac{9}{25}+\frac{16}{25}=\frac{9 + 16}{25}=\frac{25}{25}=1$.

Answer:

We need to show that the point satisfies the equation of the unit circle, that is, $x^{2}+y^{2}=1$.
$x^{2}+y^{2}=(\frac{3}{5})^{2}+(-\frac{4}{5})^{2}$
$=\frac{9}{25}+\frac{16}{25}$
$=1$
Hence, the point is on the unit circle.