Find the slope of the secant line passing through

Find the slope of the secant line passing through the points $M_1(1,\,y_1)$ and $M_2(2,\,y_2)$ on the hyperbola $y = \frac{1}{x}$.

Step 1: Determine the $y$-coordinates on the hyperbola

Both points lie on the curve $y = \dfrac{1}{x}$, so we substitute the given $x$-values: \[y_1 = \frac{1}{1} = 1, \qquad y_2 = \frac{1}{2}.\]

The two points are therefore

\[ M_1 = (1,\;1), \qquad M_2 = \left(2,\;\tfrac{1}{2}\right). \]

Step 2: Apply the slope formula

The slope $k$ of the line through $M_1(x_1, y_1)$ and $M_2(x_2, y_2)$ is

\[ k = \frac{y_2 - y_1}{x_2 - x_1}. \]

Substituting the coordinates: \[k = \frac{\dfrac{1}{2} - 1}{2 - 1} = \frac{\dfrac{1}{2} - \dfrac{2}{2}}{1} = \frac{-\dfrac{1}{2}}{1} = -\frac{1}{2}.\]

Step 3: Verification

The secant line through $M_1(1,1)$ with slope $k = -\tfrac{1}{2}$ is

\[y - 1 = -\frac{1}{2}(x - 1) \;\Longrightarrow\; y = -\frac{x}{2} + \frac{3}{2}\]

\[x = 2:\quad y = -\frac{2}{2} + \frac{3}{2} = -1 + \frac{3}{2} = \frac{1}{2} \;\;\checkmark\]

The line does pass through $M_2\left(2,\tfrac12\right)$, which confirms the slope.