Find the equation of all lines having slope

Solution:

The equation of the given curve is $y=\frac{1}{x-1}, x \neq 1$.

The slope of the tangents to the given curve at any point (x, y) is given by,

$\frac{d y}{d x}=\frac{-1}{(x-1)^{2}}$

If the slope of the tangent is −1, then we have:

$\frac{-1}{(x-1)^{2}}=-1$

$\Rightarrow(x-1)^{2}=1$

$\Rightarrow x-1=\pm 1$

$\Rightarrow x=2,0$

When x = 0, y = −1 and when x = 2, y = 1.

Thus, there are two tangents to the given curve having slope −1. These are passing through the points (0, −1) and (2, 1).

∴The equation of the tangent through (0, −1) is given by,

$y-(-1)=-1(x-0)$

$\Rightarrow y+1=-x$

$\Rightarrow y+x+1=0$

∴The equation of the tangent through (2, 1) is given by,

$y-1=-1(x-2)$

$\Rightarrow y-1=-x+2$

$\Rightarrow y+x-3=0$

Hence, the equations of the required lines are y + x + 1 = 0 and y + x − 3 = 0.