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

Answers (1)

We know that the slope of the tangent of at the point of the given curve is given by  \frac{dy}{dx}

Given the equation of curve is
y = \frac{1}{x-1}
\frac{dy}{dx} = \frac{-1}{(1-x)^2}
It is given thta slope is -1
So,
\frac{-1}{(1-x)^2} = -1 \Rightarrow (1-x)^2 = 1 = 1 - x = \pm 1 \\ \\ x = 0 \ and \ x = 2
Now, when x = 0 , y = \frac{1}{x-1} = \frac{1}{0-1} = -1
and 
when x = 2 , y = \frac{1}{x-1} = \frac{1}{(2-1)} = 1
Hence, the coordinates are (0,-1) and (2,1)
Equation of line passing through (0,-1) and having slope = -1 is
y = mx + c
-1 = 0 X -1 + c
c = -1
Now equation of line is 
y = -x -1
y + x + 1 = 0
Similarly, Equation of line passing through (2,1) and having slope = -1 is
y = mx  + c
1 = -1X2 + c
c = 3
Now equation of line is 
y = -x + 3
y + x - 3 = 0

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