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It is given that y = 1/(x - 1) , x ≠ 1

differentiate y with respect to x,

dy/dx = -1/(x - 1)² , x ≠ 1

so, SLOPE of tangent of the curve at (x,y) = dy/dx

= -1/(x - 1)²

A/C to question,

we equation of all lines having slope -1 that are tangents to the curve, y = 1/(x -1) , x ≠ 1.

so, slope of tangent = - 1

=> -1/(x - 1)² = -1

=> (x - 1)² = 1

square root both SIDES,

x - 1 = ± 1,

x = 1 ± 1 = 0, 2

put x = 0, in the curve y = 1/(x - 1)

y = 1/(0 - 1) = -1

now equation of line PASSING through (0,-1) :

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

=> y + 1 + x = 0

hence, equation is x + y + 1 = 0

x+y+1=0

put x = 2, in the curve y = 1/(x - 1)

y = 1/(2 - 1) = 1

now equation of line passing through (2,1) :

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

=> y - 1 + x - 2 = 0

=> x + y - 3 = 0

hence, equation is x + y - 3 = 0

x+y−3=0