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# Find the equation of all lines having slope $–1$ that are tangents to the curve $y= \large\frac{1}{x-1}, $$\: x \neq 1 \begin{array}{1 1} (A)\;x+y+1=0 ; x+y-3=0 \\ (B)\;x+y-1=0 ; x+y-3=0 \\ (C)\;x+y+1=0 ; x-y-3=0 \\ (D)\;x-y-1=0 ; x+y-3=0 \end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • If y=f(x),then \big(\large\frac{dy}{dx}\big)_P=slope of the tangent to y=f(x) at the point P. • Equation of line with given two points is \large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1} Step 1: Given :y=\large\frac{1}{x-1} Differentiating with respect to x we get, \large\frac{dy}{dx}=\frac{-1}{(x-1)^2}-----(1) The given slope is -1 Therefore \large\frac{dy}{dx}$$=-1$----(2)
Step 2:
Now equating equ(1) and (2) we get,

+1 vote