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My book had written that slope of tangent line is

$$m=\frac{dy}{dx}$$ And, slope of normal is $$-\frac{1}{m}=-\frac{dx}{dy}$$

It was little bit weird when I was solving problems. They had found that slope of tangent line is (0,7)

$$m_1=\frac{1}{20}$$

Then, when they wrote slope of normal it wasn't as above equation they had changed something I guess. $$m_2=-20$$

That's what they wrote. But, if I put values in that equation than, I get

$$-\frac{1}{m}=-20$$ $$\frac{1}{m}=20$$ $$m=\frac{1}{20}$$ It is as slope of tangent. But, how they found $-20$? It is in Cartesian Coordinate system.

Moni145
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1 Answers1

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The slope of the perpendicular line is the negative reciprocal of the slope of the tangent line. You don't need to solve for $m$ for anything. When you do this:

$\frac{-1}{m} = -20 \implies m=\frac{1}{20}$

You've just gone backwards, and asked the question "what would the tangent line slope be where the perpendicular slope is $-20$?" You're thinking about it too hard.

If the slope of the tangent line at a point is $\frac{1}{20}$, then the slope of the perpendicular line is the negative reciprocal of $\frac{1}{20}$, or $-20$.

Moni145
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    $-20$ how? Which equation was applied? negative negative goes way. But, it didn't remove in main equation also.. Why? –  Jul 16 '21 at 09:30
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    If the slope of the tangent line is $m_{tan}$, then the slope of the perpendicular line at that point is $\frac{-1}{m_{tan}}$. In other words, the slope of the perpendicular line is the opposite reciprocal of the slope of the tangent line. If $m_{tan} = \frac{1}{20},$ then the opposite reciprocal of it, or $\frac{-1}{m_{tan}}$, is $\frac{-1}{\frac{1}{20}}$, or $\frac{-1}{1}\times\frac{20}{1}$, or $-20$. – Moni145 Jul 16 '21 at 09:33