Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\overrightarrow {AB} = \left( { - 3; - 2} \right)\\
\overrightarrow {AC} = \left( { - 5; - 3} \right)\\
\overrightarrow {BC} = \left( { - 3; - 1} \right)\\
b,\\
AB = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 2} \right)}^2}} = \sqrt {13} \\
AC = \sqrt {{{\left( { - 5} \right)}^2} + {{\left( { - 3} \right)}^2}} = \sqrt {34} \\
BC = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 1} \right)}^2}} = \sqrt {10} \\
c,\\
{m_a}^2 = \frac{{A{B^2} + A{C^2}}}{2} - \frac{{B{C^2}}}{4}\\
\Leftrightarrow {m_a}^2 = \frac{{13 + 34}}{2} - \frac{{10}}{4}\\
\Leftrightarrow {m_a}^2 = 21\\
\Leftrightarrow {m_a} = \sqrt {21} \\
d,\\
\cos B = \frac{{B{A^2} + B{C^2} - A{C^2}}}{{2BA.BC}} = \frac{{10 + 13 - 34}}{{2.\sqrt {10} .\sqrt {13} }} < 0\\
\Rightarrow 90^\circ < \widehat B < 180^\circ \\
e,\\
p = \frac{{AB + BC + CA}}{2} = \frac{{\sqrt {10} + \sqrt {13} + \sqrt {34} }}{2}\\
\Rightarrow {S_{ABC}} = \sqrt {p\left( {p - AB} \right)\left( {p - AC} \right)\left( {p - BC} \right)} = \frac{{\sqrt {399} }}{4}
\end{array}\)
