Đáp án:
d) E(-7;0)
Giải thích các bước giải:
\(\begin{array}{l}
a)\overrightarrow {AB} = \left( {2;3} \right) \to AB = \sqrt {13} \\
\overrightarrow {AC} = \left( { - 6;5} \right) \to AC = \sqrt {61} \\
\overrightarrow {BC} = \left( { - 8; - 2} \right) \to BC = 2\sqrt {17} \\
{P_{ABC}} = \sqrt {13} + \sqrt {61} + 2\sqrt {17} \\
b)Gs:G\left( {x;y} \right)\\
\to \left\{ \begin{array}{l}
x = \dfrac{{1 + 3 - 5}}{3}\\
y = \dfrac{{ - 2 + 1 + 3}}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - \dfrac{1}{3}\\
y = \dfrac{2}{3}
\end{array} \right.\\
\to G\left( { - \dfrac{1}{3};\dfrac{2}{3}} \right)\\
c)Do:D \in Oy\\
\to D\left( {0;y} \right)\\
\overrightarrow {AB} = \left( {2;3} \right) \to AB = \sqrt {13} \to A{B^2} = 13\\
\overrightarrow {AD} = \left( { - 1;y + 2} \right) \to A{D^2} = 1 + {y^2} + 4y + 4
\end{array}\)
Do ΔABD cân A
\(\begin{array}{l}
\to A{B^2} = A{D^2}\\
\to 13 = {y^2} + 4y + 5\\
\to {y^2} + 4y - 8 = 0\\
\to \left[ \begin{array}{l}
y = - 2 + 2\sqrt 3 \\
y = - 2 - 2\sqrt 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
D\left( {0; - 2 + 2\sqrt 3 } \right)\\
D\left( {0; - 2 - 2\sqrt 3 } \right)
\end{array} \right.\\
d)Gs:E\left( {a;b} \right)\\
\overrightarrow {AB} = \left( {2;3} \right)\\
\overrightarrow {EC} = \left( { - 5 - a;3 - b} \right)
\end{array}\)
Do ABCE là hình bình hành
\(\begin{array}{l}
\to \overrightarrow {AB} = \overrightarrow {EC} \\
\to \left\{ \begin{array}{l}
2 = - 5 - a\\
3 = 3 - b
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a = - 7\\
b = 0
\end{array} \right.\\
\to E\left( { - 7;0} \right)
\end{array}\)
