Measure altitude

As below:

Triangle RST has vertices R (-3,4), S (-1,0), T (3,2). Find the length of the altitude from S to RT.

$\boxed{~ \begin{array}{rcl} \vec{RT} &=& \vec{R} - \vec{T} = \binom{-3}{4} - \binom{3}{2} = \binom{-6}{2}\\\\ \overline{RT}=|\vec{RT}| &=&\sqrt{ \vec{RT}\cdot \vec{RT} }\\ \overline{RT} &=& \sqrt{ \binom{-6}{2} \binom{-6}{2} }\\ \overline{RT} &=& \sqrt{36+4 }\\ \overline{RT} &=& \sqrt{40}\\ \overline{RT} &=& \sqrt{4\cdot 10}\\ \mathbf{\overline{RT}}& \mathbf{=} & \mathbf{2\sqrt{10} }\\ \end{array} ~}$

$\boxed{~ \begin{array}{rcl} \vec{ST} &=& \vec{S} - \vec{T} = \binom{-1}{0} - \binom{3}{2} = \binom{-4}{-2}\\\\ \overline{RT} \cdot h &=& |\vec{RT} \times \vec{ST}|\\ h &=& \frac{|\vec{RT} \times \vec{ST}|}{\overline{RT}}\\ h &=& \frac{ \left| \binom{-6}{2} \times\binom{-4}{-2} \right| }{2\sqrt{10}}\\ h &=& \frac{(-6)\cdot (-2) - (2) \cdot (-4) }{2\sqrt{10}}\\ h &=& \frac{ 12 + 8 }{2\sqrt{10}}\\ h &=& \frac{ 20 }{2\sqrt{10}}\\ h &=& \frac{ 10 }{ \sqrt{10} } \cdot \frac{ \sqrt{10} }{ \sqrt{10} }\\ h &=& \frac{ 10 }{ 10 } \cdot \sqrt{10} \\ \mathbf{h} &\mathbf{=}& \mathbf{\sqrt{10}} \\ \end{array} ~}$