What makes slope rise over run? What makes the standard equation for a line use a slope of rise over run as opposed to run over rise?
What would the standard equation of a line look like if m was equal to run/rise $\frac{\Delta x}{\Delta y}$
[Math] Why is slope rise/run
algebra-precalculus
Related Solutions
If the length of the pendulum string is $L$ and the string makes an angle $\theta$ with the vertical, then the vertical distance (from anchor point of the string down the the height of the bob) is $$ L \cos \theta. $$ (Do you see why?)
Now, the height of the bob, measured up from the lowest point is $$ h = L - L \cos \theta = L(1 - \cos \theta). $$
Now, plug in $L = 1.8$m and $\theta = 24^{\circ}$ to find $h$.
The equation of the line is $$y=ax+b$$ Assume we know that $(x_1,y_1)$ and $(x_2,y_2)$ are two points on the line. Then the following system holds $$\begin{cases}y_1=ax_1+b\\y_2=ax_2+b\end{cases}$$ Now, this is a system of two equations in two unknowns ($a$ and $b$). Solving it (by subtracting the first equation from the second), yields $$a=\frac{y_2-y_1}{x_2-x_1}$$
Graphically
Best Answer
If you write the equation of a line as $$y = mx+b$$ then it comes from this naturally. That's because if you move along the line a little bit from the point $(x,y)$ to the point $(x+\Delta x, y+\Delta y)$, then the new point satisfies the equation too. So you have that $$y+\Delta y = m(x+\Delta x) + b = mx + m\,\Delta x + b$$ Subtracting these gives $$\Delta y = m\,\Delta x$$ which means that $$\frac{\textrm{rise}}{\textrm{run}} =\dfrac{\Delta y}{\Delta x} = m$$ You can call it what you want, but the coefficient of $x$ in the original equation is the rise over the run. If you want to call run/rise the slope, then its reciprocal would be the coefficient of $x$.