# Finding the slope of a function

How do I find the slope of this function:

$px + (2p-1)y + 4 = 0$

I need to know how to answer a previous question of mine (also posted on this forum)

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You should first rearrange so that $y$ is on the left hand side:

$$(2p-1)y = px - 4$$

Then divide through by $2p-1$:

$$y = \frac{-p}{2p-1} x - \frac{4}{2p-1}$$

The slope of the line is the coefficient of $x$ in this equation.

Note that this only works if $2p-1\neq 0$ (because you can't divide by zero), which means that you need $p\neq 1/2$.

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I think that how this a line, is only, \begin{eqnarray} px + (2p-1)y+4=0 &\Leftrightarrow& y = \dfrac{-p}{(2p-1)}x + \dfrac{4}{(2p-1)}. \end{eqnarray} Then the slop is $\dfrac{-p}{(2p-1)}$, isn't it?

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seems so, I would mention $p\neq 0.5$ –  Belgi Oct 4 '12 at 17:01
change 4 to -4 ;) –  Alex Oct 4 '12 at 17:04

For a straight line given by equation $Ax+By+C=0$ the slope coefficient equals $-\frac{A}{B}$

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Do you know scalar- or inner product?

The equation is $\langle (p,2p-1); (x,y)\rangle = -4$, and this is parallel to the plain containing the origo $\{(x,y)\in\mathbb R^2 \mid \langle (p,2p-1); (x,y)\rangle = 0\}$. And satisfying this equation means that the vector with coordinates $(x,y)$ is perpendicular to $(p,2p-1)$, so this latter is the normalvector of the line..

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you could differentiate it:

$$px + (2p-1)y + 4 = 0$$ $$p + (2p-1)\frac{dy}{dx} = 0$$ giving: $$\frac{dy}{dx}=-\frac{p}{2p-1}$$

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By the definition of slope,you need to find the value of $\frac{dy}{dx}$ for your curve.To find that out,just differentiate both sides with respect to $x$,then we'll get:
$p+(2p-1)y'=0$
It can be solved for $y'$. Important thing though is if $p$ is a function rather than a constant,then while differentiating,we'll have to use chain rules,etc, and proceed similarly.