Each of the given coordinates lie on the terminal arm of an angle $\theta$ in standard position. Determine the trigonometric ratio in simplified radical form for each coordinate.
$(-4,6)$
$(4,6)$
$(-6,-4)$
$(6,4)$

Step 1 ：First, we need to understand that the coordinates given are points on the terminal arm of an angle $\theta$ in standard position. This means that the coordinates represent the end point of the line segment that forms the angle $\theta$ with the positive x-axis.

Step 2 ：The trigonometric ratios we are interested in are sine, cosine and tangent. These are defined as follows for an angle $\theta$ in standard position with a point $(x,y)$ on its terminal arm: $\sin(\theta) = \frac{y}{r}$, $\cos(\theta) = \frac{x}{r}$, and $\tan(\theta) = \frac{y}{x}$, where $r$ is the distance from the origin to the point, given by $r = \sqrt{x^2 + y^2}$.

Step 3 ：Let's start with the point $(-4,6)$. The distance from the origin to this point is $r = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$. So, the trigonometric ratios are $\sin(\theta) = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}}$, $\cos(\theta) = \frac{-4}{2\sqrt{13}} = -\frac{2}{\sqrt{13}}$, and $\tan(\theta) = \frac{6}{-4} = -\frac{3}{2}$.

Step 4 ：Next, consider the point $(4,6)$. The distance from the origin to this point is $r = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$. So, the trigonometric ratios are $\sin(\theta) = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}}$, $\cos(\theta) = \frac{4}{2\sqrt{13}} = \frac{2}{\sqrt{13}}$, and $\tan(\theta) = \frac{6}{4} = \frac{3}{2}$.

Step 5 ：For the point $(-6,-4)$, the distance from the origin is $r = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$. So, the trigonometric ratios are $\sin(\theta) = \frac{-4}{2\sqrt{13}} = -\frac{2}{\sqrt{13}}$, $\cos(\theta) = \frac{-6}{2\sqrt{13}} = -\frac{3}{\sqrt{13}}$, and $\tan(\theta) = \frac{-4}{-6} = \frac{2}{3}$.

Step 6 ：Finally, for the point $(6,4)$, the distance from the origin is $r = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$. So, the trigonometric ratios are $\sin(\theta) = \frac{4}{2\sqrt{13}} = \frac{2}{\sqrt{13}}$, $\cos(\theta) = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}}$, and $\tan(\theta) = \frac{4}{6} = \frac{2}{3}$.

Step 7 ：In conclusion, the trigonometric ratios for the points $(-4,6)$, $(4,6)$, $(-6,-4)$, and $(6,4)$ are $\boxed{\left(-\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}, -\frac{3}{2}\right)}$, $\boxed{\left(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}, \frac{3}{2}\right)}$, $\boxed{\left(-\frac{3}{\sqrt{13}}, -\frac{2}{\sqrt{13}}, \frac{2}{3}\right)}$, and $\boxed{\left(\frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}}, \frac{2}{3}\right)}$ respectively.