using the point-slope form of the line: y - y_1 = m(x - x_1) to find the equation of the line passing through the points (5, -3) and (-1, 5)
First find the slope:
$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-(-3)}{-1-5}=\frac{8}{-6}=-\frac{4}{3}$.
Now plug $\displaystyle m$ and either point into $\displaystyle y-y_0=m\left(x-x_0\right)$ and that will give you the line you're looking for.
Can you try to finish this?
See the PurpleMath article, you need the second worked example.
CB
Oooh, I'd like to answer this as we just learned it yesterday .
To get the equation of a line passing through two points, you need to know how to work out the gradient:
$\displaystyle m = \frac {y_2 - y_1}{x_2 - x_1}$
or:
$\displaystyle m = \frac {y_1 - y_2}{x_1 - x_2}$
So to work that out with your question:
$\displaystyle m = \frac {y_1 - y_2}{x_2 - x_1} = \frac{-3 - 5}{5 - (-1)} = \frac{-8}{6} = -\frac{4}{3}$
So you now have the gradient, $\displaystyle -\frac{4}{3}$. Now choose one of the coordinates, I'll use $\displaystyle (5, -3)$.
$\displaystyle \begin{array}{rcrcrc}
y - y_1 = m(x - x_1)\\
y - (-3) = -\frac{4}{3}(x - 5)\\
3(y + 3) = -4(x - 5)\\
3y + 9 = -4x + 20\\
3y = -4x + 11\\
y = \frac{-4x + 11}{3}
\end{array}$
I've checked that against both of your coordinates and it works, so look over those notes and you should be able to understand it .