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# Find the equations of all lines having slope 2 which are tangent to the curve y = 1/(x - 3), x ≠ 3

**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 slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

The given curve is y

= 1 / (x - 3)

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

dy/dx

= d/dx 1/(x - 3)

- 1 / (x - 3)^{2}

The slope of the tangent is 2.

So, we have:

- 1 / (x - 3)^{2} = 2

⇒ 2 (x - 3)^{2} = - 1

⇒ (x - 3)^{2}

= - 1/2

It is not possible since the LHS is positive and RHS is negative.

Thus, there is no tangent to the curve of slope 2

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 11

## Find the equations of all lines having slope 2 which are tangent to the curve y = 1/(x - 3), x ≠ 3.

**Summary:**

There is no equation possible for all lines having slope 2 which are tangent to the curve y = 1/(x - 3), x ≠ 3

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