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.
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:
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 $-20$.