a) Ta có: \(\overrightarrow {AB} = \left( { - 28;6} \right),\overrightarrow {BC} = \left( {1; - 9} \right),\overrightarrow {AC} = \left( { - 27; - 3} \right)\)
\( \Rightarrow \overrightarrow {AC} .\overrightarrow {BC} = \left( { - 27} \right).1 + \left( { - 9} \right).\left( { - 3} \right) = 0\)
\( \Rightarrow AC \bot BC\) hay tam giác ABC vuông tại C.
b) Gọi D(x;y), ta có \(\overrightarrow {AD} = \overrightarrow {BC} \Leftrightarrow \left\{ \begin{array}{l}x - 34 = 1\\y + 1 = - 9\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 35\\y = - 10\end{array} \right. \Rightarrow D\left( {35; - 10} \right)\)
c) Gọi \(E\left( {x;0} \right) \in Ox\), ta có \(\overrightarrow {AE} = \left( {x - 34;1} \right),\overrightarrow {BE} = \left( {x - 6; - 5} \right)\)
\(AE = BE \Leftrightarrow \sqrt {{{\left( {x - 34} \right)}^2} + {1^2}} = \sqrt {{{\left( {x - 6} \right)}^2} + {{\left( { - 5} \right)}^2}} \)
\( \Leftrightarrow {x^2} - 68x + 1156 + 1 = {x^2} - 12x + 36 + 25\)
\( \Leftrightarrow - 56x + 1096 = 0 \Leftrightarrow x = \dfrac{{137}}{7}\)
Vậy \(E\left( {\dfrac{{137}}{7};0} \right)\).
