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

**Solution:**

For a curve y = f(x) containing the point (x_{1},y_{1}) the equation of the tangent line to the curve at (x_{1},y_{1}) is given by

y − y_{1} = f′(x_{1}) (x − x_{1}).

The given curve is

y = 1/(1 - x)

The slope of the tangents to the given curve is given by,

dy/dx

= d/dx (1/(1 - x))

= - 1/(1 - x)^{2}

Given that the slope of the tangent is - 1

So, we have:

- 1/(1 - x)^{2} = - 1

⇒ ( x - 1)^{2} = 1

⇒ x - 1 = ± 1

⇒ 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 and passing through the points (0, - 1) and (2, 1).

Hence,

the equation of the tangent through (0, - 1) is given by,

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

⇒ y + 1 = - x

⇒ y + x + 1 = 0

The equation of the tangent through (2, 1)

y - 1 = - 1(x - 2) is given by,

⇒ y - 1 = x + 2

⇒ y + x - 3 = 0

Thus, the equations of the required lines are y + x + 1 = 0 and y + x - 3 = 0

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 10

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

**Summary:**

The equation of all lines having slope - 1 that are tangents to the curve y = 1/(1 - x), x ≠ 1 are y + x + 1 = 0 and y + x - 3 = 0

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