Toán Lớp 9: Cho 3 đường thẳng
y = mx + 2 ( d1)
y = x + 2m ( d2 )
y = -3x + 4 ( d3 )
tìm m để d1 , d2 , d3 đồng quy
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Toán Lớp 9: Cho 3 đường thẳng
y = mx + 2 ( d1)
y = x + 2m ( d2 )
y = -3x + 4 ( d3 )
tìm m để d1 , d2 , d3 đồng quy
Comments ( 1 )
m = \dfrac{{ – 1 + \sqrt {41} }}{2}\\
m = \dfrac{{ – 1 – \sqrt {41} }}{2}
\end{array} \right.\)
\left\{ \begin{array}{l}
x + 2m = y\\
– 3x + 4 = y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4x + 2m – 4 = 0\\
– 3x + 4 = y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m – 2}}{2}\\
– 3.\dfrac{{m – 2}}{2} + 4 = y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m – 2}}{2}\\
y = \dfrac{{ – 3m + 6 + 8}}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m – 2}}{2}\\
y = \dfrac{{ – 3m + 14}}{2}
\end{array} \right.
\end{array}\)
\to Thay:x = \dfrac{{m – 2}}{2};y = \dfrac{{ – 3m + 14}}{2}\\
\to \left( {{d_1}} \right):m.\dfrac{{m – 2}}{2} + 2 = \dfrac{{ – 3m + 14}}{2}\\
\to {m^2} – 2m + 4 = – 3m + 14\\
\to {m^2} + m – 10 = 0\\
\to \left[ \begin{array}{l}
m = \dfrac{{ – 1 + \sqrt {41} }}{2}\\
m = \dfrac{{ – 1 – \sqrt {41} }}{2}
\end{array} \right.
\end{array}\)