Cho điểm A(2;-3), B(4;3), C(-2;4). Tính D(x;y) sao cho tam giác ABD vuông căn tại D. Giúp vs ạ ! 07/08/2021 Bởi Arya Cho điểm A(2;-3), B(4;3), C(-2;4). Tính D(x;y) sao cho tam giác ABD vuông căn tại D. Giúp vs ạ !
Đáp án: \[\left[ \begin{array}{l}D\left( {0;1} \right)\\D\left( {\frac{{28}}{5}; – \frac{9}{5}} \right)\end{array} \right.\] Giải thích các bước giải: Tam giác ABD vuông cân tại D khi và chỉ khi \(\left\{ \begin{array}{l}AD \bot DB\\AD = DB\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\overrightarrow {AD} .\overrightarrow {BD} = 0\\AD = DB\end{array} \right.\) Ta có: \(\begin{array}{l}D\left( {x;y} \right) \Rightarrow \left\{ \begin{array}{l}\overrightarrow {AD} \left( {x – 2;y + 3} \right)\\\overrightarrow {BD} \left( {x – 4;y – 3} \right)\end{array} \right. \Rightarrow \left\{ \begin{array}{l}AD = \sqrt {{{\left( {x – 2} \right)}^2} + {{\left( {y + 3} \right)}^2}} \\BD = \sqrt {{{\left( {x – 4} \right)}^2} + {{\left( {y – 3} \right)}^2}} \end{array} \right.\\\left\{ \begin{array}{l}\overrightarrow {AD} .\overrightarrow {BD} = 0\\AD = DB\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left( {x – 2} \right)\left( {x – 4} \right) + \left( {y + 3} \right)\left( {y – 3} \right) = 0\\\sqrt {{{\left( {x – 2} \right)}^2} + {{\left( {y + 3} \right)}^2}} = \sqrt {{{\left( {x – 4} \right)}^2} + {{\left( {y – 3} \right)}^2}} \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{x^2} – 6x + 8 + {y^2} – 9 = 0\\ – 2x + 4 + 6y + 9 = – 8x + 16 – 6y + 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{x^2} – 6x + {y^2} = 1\\6x + 12y = 12\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}{x^2} – 6x + {y^2} = 1\\x + 2y = 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 0\\y = 1\end{array} \right.\\\left\{ \begin{array}{l}x = \frac{{28}}{5}\\y = – \frac{9}{5}\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}D\left( {0;1} \right)\\D\left( {\frac{{28}}{5}; – \frac{9}{5}} \right)\end{array} \right.\end{array}\) Bình luận
Đáp án:
\[\left[ \begin{array}{l}
D\left( {0;1} \right)\\
D\left( {\frac{{28}}{5}; – \frac{9}{5}} \right)
\end{array} \right.\]
Giải thích các bước giải:
Tam giác ABD vuông cân tại D khi và chỉ khi \(\left\{ \begin{array}{l}
AD \bot DB\\
AD = DB
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\overrightarrow {AD} .\overrightarrow {BD} = 0\\
AD = DB
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
D\left( {x;y} \right) \Rightarrow \left\{ \begin{array}{l}
\overrightarrow {AD} \left( {x – 2;y + 3} \right)\\
\overrightarrow {BD} \left( {x – 4;y – 3} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
AD = \sqrt {{{\left( {x – 2} \right)}^2} + {{\left( {y + 3} \right)}^2}} \\
BD = \sqrt {{{\left( {x – 4} \right)}^2} + {{\left( {y – 3} \right)}^2}}
\end{array} \right.\\
\left\{ \begin{array}{l}
\overrightarrow {AD} .\overrightarrow {BD} = 0\\
AD = DB
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left( {x – 2} \right)\left( {x – 4} \right) + \left( {y + 3} \right)\left( {y – 3} \right) = 0\\
\sqrt {{{\left( {x – 2} \right)}^2} + {{\left( {y + 3} \right)}^2}} = \sqrt {{{\left( {x – 4} \right)}^2} + {{\left( {y – 3} \right)}^2}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} – 6x + 8 + {y^2} – 9 = 0\\
– 2x + 4 + 6y + 9 = – 8x + 16 – 6y + 9
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} – 6x + {y^2} = 1\\
6x + 12y = 12
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} – 6x + {y^2} = 1\\
x + 2y = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 0\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = \frac{{28}}{5}\\
y = – \frac{9}{5}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
D\left( {0;1} \right)\\
D\left( {\frac{{28}}{5}; – \frac{9}{5}} \right)
\end{array} \right.
\end{array}\)