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Equal segments

Let f(x) = \frac{1}{x} and let (\varepsilon) be a line that intersects \mathcal{C}_f at the points \mathrm{A} , \Delta and the axis at the points \mathrm{B} and \Gamma.

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Prove that \mathrm{AB}= \Gamma \Delta.

Solution

Let (\varepsilon):y= \alpha x + \beta with \alpha \neq 0. Then,

    \[\alpha x + \beta = \frac{1}{x} \Leftrightarrow \alpha x^2 + \beta x - 1 = 0 \Leftrightarrow \left\{\begin{matrix} x_1 + x_2 &= & - \dfrac{\beta}{\alpha} \\\\ x_1 x_2 & = & - \dfrac{1}{\alpha} \end{matrix}\right.\]

Hence,

    \begin{align*} \overrightarrow{\mathrm{BA}} &= \left ( x_1, \frac{1}{x_1} - \beta \right ) \\ &= \left ( - \frac{\beta}{\alpha} -x_2, - \frac{1}{x_2} \right ) \\ &= \overrightarrow{\Delta \Gamma} \end{align*}

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