Toán Lớp 10: Viết pt đg rron (C) nhận AB là đg Kính Biết
a) A(1;1) B(9,2)
b) A(3,5) B(9,2)
Cần trợ giúp ạ
5sao
Ạ cmon
C) A(-3,4) B(1,-1)
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Toán Lớp 10: Viết pt đg rron (C) nhận AB là đg Kính Biết
a) A(1;1) B(9,2)
b) A(3,5) B(9,2)
Cần trợ giúp ạ
5sao
Ạ cmon
C) A(-3,4) B(1,-1)
Comments ( 1 )
\left\{ \begin{array}{l}
{x_I} = \dfrac{{{x_A} + {x_B}}}{2} = \dfrac{{3 + 9}}{2} = 6\\
{y_I} = \dfrac{{{y_A} + {y_B}}}{2} = \dfrac{{5 + 2}}{2} = \dfrac{7}{2}
\end{array} \right.\\
\Rightarrow I\left( {6;\dfrac{7}{2}} \right)\\
R = \dfrac{{AB}}{2} = \dfrac{{\sqrt {{{\left( {9 – 3} \right)}^2} + {{\left( {5 – 2} \right)}^2}} }}{2} = \dfrac{{3\sqrt 5 }}{2}\\
\left( C \right):{\left( {x – 6} \right)^2} + {\left( {y – \dfrac{7}{2}} \right)^2} = \dfrac{{45}}{4}
\end{array}$
\left\{ \begin{array}{l}
{x_I} = \dfrac{{{x_A} + {x_B}}}{2} = \dfrac{{1 – 3}}{2} = – 1\\
{y_I} = \dfrac{{{y_A} + {y_B}}}{2} = \dfrac{{4 – 1}}{2} = \dfrac{3}{2}
\end{array} \right.\\
\Rightarrow I\left( { – 1;\dfrac{3}{2}} \right)\\
R = \dfrac{{AB}}{2} = \dfrac{{\sqrt {{{\left( {1 + 3} \right)}^2} + {{\left( {4 + 1} \right)}^2}} }}{2} = \dfrac{{\sqrt {41} }}{2}\\
\left( C \right):{\left( {x + 1} \right)^2} + {\left( {y – \dfrac{3}{2}} \right)^2} = \dfrac{{41}}{4}
\end{array}$