Đáp án:
a.\(M(18;4)\)
b. \(\left[ \begin{array}{l}
C(6;0)\\
C(0;0)
\end{array} \right.\)
c. \(D( - 10;0)\)
Giải thích các bước giải:
a. Gs M(x;y)
\(\begin{array}{l}
\overrightarrow {MA} = ( - 2 - x;4 - y)\\
\overrightarrow {MB} = (8 - x;4 - y)
\end{array}\)
Có:
\(\overrightarrow {MA} = 2\overrightarrow {MB} \to \left\{ \begin{array}{l}
- 2 - x = 16 - 2x\\
4 - y = 8 - 2y
\end{array} \right. \to \left\{ \begin{array}{l}
x = 18\\
y = 4
\end{array} \right. \to M(18;4)\)
b. Do C∈Ox⇒C(a;0)
\(\begin{array}{l}
\overrightarrow {AC} = (a + 2; - 4)\\
\overrightarrow {BC} = (a - 8; - 4)
\end{array}\)
Do ΔABC vuông tại C ⇒\(\overrightarrow {AC} .\overrightarrow {BC} = \ 0 \)
\( \to {a^2} - 6a - 16 + 16 = 0 \to \left[ \begin{array}{l}
a = 6\\
a = 0
\end{array} \right. \to \left[ \begin{array}{l}
C(6;0)\\
C(0;0)
\end{array} \right.\)
c. Gs D(m;n)
\(\begin{array}{l}
\overrightarrow {AB} = (10;0)\\
\overrightarrow {DO} = ( - m; - n)
\end{array}\)
Do ABOD là hbh
\( \to \overrightarrow {AB} = \overrightarrow {DO} \to \left\{ \begin{array}{l}
m = - 10\\
n = 0
\end{array} \right. \to D( - 10;0)\)
