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Question

(A) The equation of the given curve is $y=\frac{1}{x-1}, x 
eq 1$. The slope of the tangent to the curve at any point $(x, y)$ is given by the deriva

Vedclass · Accepted Answer

$y+x+1=0$ and $y+x-3=0$

Vedclass · Answer

(A) The equation of the given curve is $y=\frac{1}{x-1}, x 
eq 1$. The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx} = \frac{-1}{(x-1)^2}$. Given that the slope of the tangent is $-1$,we set the derivative equal to $-1$: $\frac{-1}{(x-1)^2} = -1$ $(x-1)^2 = 1$ $x-1 = \pm 1$ $x = 2$ or $x = 0$. For $x=0$,$y = \frac{1}{0-1} = -1$. The point is $(0, -1)$. For $x=2$,$y = \frac{1}{2-1} = 1$. The point is $(2, 1)$. The equation of the tangent at $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$. For point $(0, -1)$ and $m = -1$: $y - (-1) = -1(x - 0) \Rightarrow y + 1 = -x \Rightarrow y + x + 1 = 0$. For point $(2, 1)$ and $m = -1$: $y - 1 = -1(x - 2) \Rightarrow y - 1 = -x + 2 \Rightarrow y + x - 3 = 0$. Thus,the equations of the required lines are $y+x+1=0$ and $y+x-3=0$.

Find the equation of all lines having slope $-1$ that are tangents to the curve $y=\frac{1}{x-1}, x \neq 1$.

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