# Find the slope of the tangent to the curve y = (x - 1)/(x - 2), x ≠ 2 at x = 10

**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 = (x - 1)/(x - 2)

Therefore,

dy/dx = d/dx (x - 1)/(x - 2)

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

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

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

Now,

the slope of the tangent to the given curve at x = 10 is given by:

dy/dx]_{x = 10} = - 1//(x - 2)^{2}]_{x = 10 }

= - 1//(10 - 2)^{2}

= - 1/64

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 2

## Find the slope of the tangent to the curve y = (x - 1)/(x - 2), x ≠ 2 at x = 10

**Summary:**

The slope of the tangent to the curve y = (x - 1)/(x - 2), x ≠ 2 at x = 10 is - 1/64. The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line

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