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Câu hỏi: Cho các số phức ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ thỏa mãn $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=2\left| {{z}_{3}} \right|=2$ và $3{{z}_{1}}{{z}_{2}}=4{{z}_{3}}\left( {{z}_{1}}+{{z}_{2}} \right)$. Gọi $A,B,C$ lần lượt là điểm biểu diễn của ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ trên mặt phẳng tọa độ. Diện tích tam giác $ABC$ bằng
A. $\dfrac{\sqrt{7}}{4}$.
B. $\dfrac{3\sqrt{7}}{4}$.
C. $\dfrac{\sqrt{7}}{2}$.
D. $\dfrac{3\sqrt{7}}{2}$.
A. $\dfrac{\sqrt{7}}{4}$.
B. $\dfrac{3\sqrt{7}}{4}$.
C. $\dfrac{\sqrt{7}}{2}$.
D. $\dfrac{3\sqrt{7}}{2}$.
Ta có : $3{{z}_{1}}{{z}_{2}}=4{{z}_{3}}\left( {{z}_{1}}+{{z}_{2}} \right)\Leftrightarrow \left| 3{{z}_{1}}{{z}_{2}} \right|=\left| 4{{z}_{3}}\left( {{z}_{1}}+{{z}_{2}} \right) \right|\Leftrightarrow \left| {{z}_{1}}+{{z}_{2}} \right|=\dfrac{3\left| {{z}_{1}} \right|.\left| {{z}_{2}} \right|}{4\left| {{z}_{3}} \right|}=\dfrac{3.2.2}{4.1}=3$
$\Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}=2{{\left| {{z}_{1}} \right|}^{2}}+2{{\left| {{z}_{2}} \right|}^{2}}-{{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{2.2}^{2}}+{{2.2}^{2}}-{{3}^{2}}=7\Leftrightarrow \left| {{z}_{1}}-{{z}_{2}} \right|=\sqrt{7}$
Đồng thời ${{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}=\left( {{z}_{1}}+{{z}_{2}} \right)\left( \overline{{{z}_{1}}+{{z}_{2}}} \right)=\left( {{z}_{1}}+{{z}_{2}} \right)\left( \overline{{{z}_{1}}}+\overline{{{z}_{2}}} \right)={{z}_{1}}.\overline{{{z}_{1}}}+{{z}_{2}}.\overline{{{z}_{2}}}+\overline{{{z}_{1}}}.{{z}_{2}}+{{z}_{1}}.\overline{{{z}_{2}}}$
$\Rightarrow \overline{{{z}_{1}}}.{{z}_{2}}+{{z}_{1}}.\overline{{{z}_{2}}}={{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}-{{\left| {{z}_{1}} \right|}^{2}}-{{\left| {{z}_{2}} \right|}^{2}}={{3}^{2}}-{{2}^{2}}-{{2}^{2}}=1$
Lại có:
$3{{z}_{1}}{{z}_{2}}=4{{z}_{3}}\left( {{z}_{1}}+{{z}_{2}} \right)\Leftrightarrow 3{{z}_{1}}\left( {{z}_{2}}-{{z}_{3}} \right)={{z}_{3}}\left( {{z}_{1}}+4{{z}_{2}} \right)\Leftrightarrow \left| {{z}_{2}}-{{z}_{3}} \right|=\dfrac{\left| {{z}_{3}} \right|.\left| {{z}_{1}}+4{{z}_{2}} \right|}{3\left| {{z}_{1}} \right|}=\dfrac{1}{6}.\left| {{z}_{1}}+4{{z}_{2}} \right|$
$3{{z}_{1}}{{z}_{2}}=4{{z}_{3}}\left( {{z}_{1}}+{{z}_{2}} \right)\Leftrightarrow 3{{z}_{2}}\left( {{z}_{1}}-{{z}_{3}} \right)={{z}_{3}}\left( 4{{z}_{1}}+{{z}_{2}} \right)\Leftrightarrow \left| {{z}_{1}}-{{z}_{3}} \right|=\dfrac{\left| {{z}_{3}} \right|.\left| 4{{z}_{1}}+{{z}_{2}} \right|}{3\left| {{z}_{2}} \right|}=\dfrac{1}{6}.\left| 4{{z}_{1}}+{{z}_{2}} \right|$
Mà $\left\{ \begin{aligned}
& {{\left| {{z}_{1}}+4{{z}_{2}} \right|}^{2}}=\left( {{z}_{1}}+4{{z}_{2}} \right)\left( \overline{{{z}_{1}}+4{{z}_{2}}} \right)=\left( {{z}_{1}}+4{{z}_{2}} \right)\left( \overline{{{z}_{1}}}+4\overline{{{z}_{2}}} \right)={{z}_{1}}.\overline{{{z}_{1}}}+16{{z}_{2}}.\overline{{{z}_{2}}}+4\overline{{{z}_{1}}}.{{z}_{2}}+4{{z}_{1}}.\overline{{{z}_{2}}}=72 \\
& {{\left| 4{{z}_{1}}+{{z}_{2}} \right|}^{2}}=\left( 4{{z}_{1}}+{{z}_{2}} \right)\left( \overline{4{{z}_{1}}+{{z}_{2}}} \right)=\left( 4{{z}_{1}}+{{z}_{2}} \right)\left( \overline{4{{z}_{1}}}+\overline{{{z}_{2}}} \right)=16{{z}_{1}}.\overline{{{z}_{1}}}+{{z}_{2}}.\overline{{{z}_{2}}}+4\overline{{{z}_{1}}}.{{z}_{2}}+4{{z}_{1}}.\overline{{{z}_{2}}}=72 \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& \left| {{z}_{2}}-{{z}_{3}} \right|=\dfrac{1}{6}.\left| {{z}_{1}}+4{{z}_{2}} \right|=\dfrac{1}{6}.\sqrt{72}=\sqrt{2} \\
& \left| {{z}_{1}}-{{z}_{3}} \right|=\dfrac{1}{6}.\left| 4{{z}_{1}}+{{z}_{2}} \right|=\dfrac{1}{6}.\sqrt{72}=\sqrt{2} \\
\end{aligned} \right.$
Khi đó $AB=\left| {{z}_{1}}-{{z}_{2}} \right|=\sqrt{7},AC=\left| {{z}_{1}}-{{z}_{3}} \right|=\sqrt{2},BC=\left| {{z}_{2}}-{{z}_{3}} \right|=\sqrt{2}$
${{S}_{ABC}}=\sqrt{p\left( p-AB \right)\left( p-AC \right)\left( p-BC \right)}=\dfrac{\sqrt{7}}{4}$.
$\Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}=2{{\left| {{z}_{1}} \right|}^{2}}+2{{\left| {{z}_{2}} \right|}^{2}}-{{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{2.2}^{2}}+{{2.2}^{2}}-{{3}^{2}}=7\Leftrightarrow \left| {{z}_{1}}-{{z}_{2}} \right|=\sqrt{7}$
Đồng thời ${{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}=\left( {{z}_{1}}+{{z}_{2}} \right)\left( \overline{{{z}_{1}}+{{z}_{2}}} \right)=\left( {{z}_{1}}+{{z}_{2}} \right)\left( \overline{{{z}_{1}}}+\overline{{{z}_{2}}} \right)={{z}_{1}}.\overline{{{z}_{1}}}+{{z}_{2}}.\overline{{{z}_{2}}}+\overline{{{z}_{1}}}.{{z}_{2}}+{{z}_{1}}.\overline{{{z}_{2}}}$
$\Rightarrow \overline{{{z}_{1}}}.{{z}_{2}}+{{z}_{1}}.\overline{{{z}_{2}}}={{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}-{{\left| {{z}_{1}} \right|}^{2}}-{{\left| {{z}_{2}} \right|}^{2}}={{3}^{2}}-{{2}^{2}}-{{2}^{2}}=1$
Lại có:
$3{{z}_{1}}{{z}_{2}}=4{{z}_{3}}\left( {{z}_{1}}+{{z}_{2}} \right)\Leftrightarrow 3{{z}_{1}}\left( {{z}_{2}}-{{z}_{3}} \right)={{z}_{3}}\left( {{z}_{1}}+4{{z}_{2}} \right)\Leftrightarrow \left| {{z}_{2}}-{{z}_{3}} \right|=\dfrac{\left| {{z}_{3}} \right|.\left| {{z}_{1}}+4{{z}_{2}} \right|}{3\left| {{z}_{1}} \right|}=\dfrac{1}{6}.\left| {{z}_{1}}+4{{z}_{2}} \right|$
$3{{z}_{1}}{{z}_{2}}=4{{z}_{3}}\left( {{z}_{1}}+{{z}_{2}} \right)\Leftrightarrow 3{{z}_{2}}\left( {{z}_{1}}-{{z}_{3}} \right)={{z}_{3}}\left( 4{{z}_{1}}+{{z}_{2}} \right)\Leftrightarrow \left| {{z}_{1}}-{{z}_{3}} \right|=\dfrac{\left| {{z}_{3}} \right|.\left| 4{{z}_{1}}+{{z}_{2}} \right|}{3\left| {{z}_{2}} \right|}=\dfrac{1}{6}.\left| 4{{z}_{1}}+{{z}_{2}} \right|$
Mà $\left\{ \begin{aligned}
& {{\left| {{z}_{1}}+4{{z}_{2}} \right|}^{2}}=\left( {{z}_{1}}+4{{z}_{2}} \right)\left( \overline{{{z}_{1}}+4{{z}_{2}}} \right)=\left( {{z}_{1}}+4{{z}_{2}} \right)\left( \overline{{{z}_{1}}}+4\overline{{{z}_{2}}} \right)={{z}_{1}}.\overline{{{z}_{1}}}+16{{z}_{2}}.\overline{{{z}_{2}}}+4\overline{{{z}_{1}}}.{{z}_{2}}+4{{z}_{1}}.\overline{{{z}_{2}}}=72 \\
& {{\left| 4{{z}_{1}}+{{z}_{2}} \right|}^{2}}=\left( 4{{z}_{1}}+{{z}_{2}} \right)\left( \overline{4{{z}_{1}}+{{z}_{2}}} \right)=\left( 4{{z}_{1}}+{{z}_{2}} \right)\left( \overline{4{{z}_{1}}}+\overline{{{z}_{2}}} \right)=16{{z}_{1}}.\overline{{{z}_{1}}}+{{z}_{2}}.\overline{{{z}_{2}}}+4\overline{{{z}_{1}}}.{{z}_{2}}+4{{z}_{1}}.\overline{{{z}_{2}}}=72 \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& \left| {{z}_{2}}-{{z}_{3}} \right|=\dfrac{1}{6}.\left| {{z}_{1}}+4{{z}_{2}} \right|=\dfrac{1}{6}.\sqrt{72}=\sqrt{2} \\
& \left| {{z}_{1}}-{{z}_{3}} \right|=\dfrac{1}{6}.\left| 4{{z}_{1}}+{{z}_{2}} \right|=\dfrac{1}{6}.\sqrt{72}=\sqrt{2} \\
\end{aligned} \right.$
Khi đó $AB=\left| {{z}_{1}}-{{z}_{2}} \right|=\sqrt{7},AC=\left| {{z}_{1}}-{{z}_{3}} \right|=\sqrt{2},BC=\left| {{z}_{2}}-{{z}_{3}} \right|=\sqrt{2}$
${{S}_{ABC}}=\sqrt{p\left( p-AB \right)\left( p-AC \right)\left( p-BC \right)}=\dfrac{\sqrt{7}}{4}$.
Đáp án A.