Đáp án:
Giải thích các bước giải:
$ A(-2;0);B(5;0)$
$\overline{AB}=\sqrt{(x_{B}-x_{A}^{2}+(y_{B}-y_{A})^{2}}=\sqrt{(5+2)^{2}+0}=7$
B)
$\begin{cases} x_{I}=\dfrac{x_{A}+x_{B}}{2}\\y_{I}=\dfrac{y_{A}+y_{B}}{2} \end{cases}$
⇒$\begin{cases} x_{I}=\dfrac{-2+5}{2}=\dfrac{3}{2}\\y_{I}=0\end{cases}$
$⇒I(\dfrac{3}{2};0)$
C)Gọi $M(x_{M};y_{M})$
$2\overrightarrow{MA}+5\overrightarrow{MB}=\overrightarrow{0}$
$⇒2(x_{A}-x_{M};y_{A}-y_{M})+5(x_{M}-y_{B};y_{B}-y_{M}=\overrightarrow{0}$
$⇒2(-2-x_{M};0-y_{M})+5(5-x_{M};0-y_{M})=\overrightarrow{0}$
⇒$\begin{cases}-4-2x_{M}+25-5x_{M}=0 \\y_{M}=0 \end{cases}$
⇒$\begin{cases}x_{M}=3 \\y_{M}=0 \end{cases}$
$⇒M(3;0)$
Gọi $N(x_{N};y_{N})$
$2\overrightarrow{NA}+3\overrightarrow{NB}=-1$
$⇒2(x_{A}-x_{M};y_{A}-y_{M})+3(x_{M}-y_{B};y_{B}-y_{M}=-1$
$⇒2(-2-x_{N};0-y_{N})+3(5-x_{n};0-y_{n})=-1$
⇒$\begin{cases}-4-2x_{N}+15-3x_{N}=-1\\y_{M}=-1\end{cases}$
⇒$\begin{cases}x_{N}=\dfrac{12}{5} \\y_{M}=-1\end{cases}$
$⇒N(\dfrac{12}{5};-1)$
