Đáp án:
\[{h_a} = \frac{1}{5}\]
Giải thích các bước giải:
\(\begin{array}{l}
A\left( {1;2} \right);\,\,\,B\left( {0;3} \right);\,\,\,\,C\left( {4;0} \right)\\
\Rightarrow \overrightarrow {AB} \left( { - 1;1} \right) \Rightarrow AB = \sqrt {{{\left( { - 1} \right)}^2} + {1^2}} = \sqrt 2 \\
\overrightarrow {BC} \left( {4; - 3} \right) \Rightarrow BC = \sqrt {{4^2} + {{\left( { - 3} \right)}^2}} = 5\\
\overrightarrow {CA} \left( { - 3;2} \right) \Rightarrow CA = \sqrt {{{\left( { - 3} \right)}^2} + {2^2}} = \sqrt {13} \\
p = \frac{{AB + BC + CA}}{2} = \frac{{5 + \sqrt 2 + \sqrt {13} }}{2}\\
{S_{ABC}} = \sqrt {p\left( {p - AB} \right).\left( {p - AC} \right).\left( {p - BC} \right)} = \frac{1}{2}\\
{h_a} = \frac{{2{S_{ABC}}}}{{BC}} = \frac{1}{5}
\end{array}\)
