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)

- Sep 18th 2009, 09:05 PMtyonnacan anyone qive me the answer and explain how to work this question for future ref.
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)

- Sep 18th 2009, 09:18 PMChris L T521
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? - Sep 18th 2009, 09:32 PMtyonna
- Sep 18th 2009, 11:49 PMmr fantastic
- Sep 19th 2009, 12:57 AMCaptainBlack
See the PurpleMath article, you need the second worked example.

CB - Sep 19th 2009, 05:24 AMViral
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 :) .