Question

The point P(2,-1) lies on the curve y=1/1- x
If Q is the point Q(x, 1/1-x), then the slope of the secant line PQ
for of x = 2.1 is​

Answer 1

SOLUTION

GIVEN

• The point P(2,-1) lies on the curve y=1/1- x

• If Q is the point Q(x, 1/1-x)

TO DETERMINE

The slope of the secant line PQ for x = 2.1

EVALUATION

Here it is given that

The point P(2,-1) lies on the curve y=1/1- x

Q is the point Q(x, 1/1-x)

So the slope of the secant line PQ in terms of PQ

$\displaystyle \sf{ = \frac{ \frac{1}{1 - x} + 1}{x - 2} }$

$\displaystyle \sf{ = \frac{ \frac{1 + 1 - x}{1 - x} }{x - 2} }$

$\displaystyle \sf{ = \frac{ \frac{2 - x}{1 - x} }{x - 2} }$

$\displaystyle \sf{ = \frac{ \frac{2 - x}{1 - x} }{ - (2 - x )} }$

$\displaystyle \sf{ = \frac{1 }{ - (1 - x )} }$

$\displaystyle \sf{ = \frac{1 }{ x - 1} }$

So For x = 2.1 the required slope

$\displaystyle \sf{ = \frac{1}{2.1 - 1} }$

$\displaystyle \sf{ = \frac{1}{1.1} }$

$\displaystyle \sf{ = \frac{10}{11} }$

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