What is the normal of x-10y=1,(-9,3) ?

The normal of x-10y=1,(-9,3) is m=(x-31)/(10(x+9))

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normal of x-10y=1,(-9,3)

normal of

x - 10 y = 1, (- 9, 3)

m = \frac{x - 31}{10 ( x + 9 )}

Solution steps

x - 10 y = 1 : y = - \frac{1 - x}{10}

Slope of the normal line at

(- 9, 3) : m = \frac{x - 31}{10 ( x + 9 )}

\frac{d ^{10}}{d x ^{10}} (x^{10})

(5 sin (x))^{^{'}}

\int_{1}^{4} 1 + \frac{9 x}{4} d x

\frac{\partial}{\partial y} (5 x^{2} y + 5 x y^{2})

\frac{d}{d x} (x (x - 1)^{2})

What is the normal of x-10y=1,(-9,3) ?
The normal of x-10y=1,(-9,3) is m=(x-31)/(10(x+9))