# How do I find the equation of the line which passes through the points (8, 1) and (-2, 3)?

Oct 4, 2014

There are several ways to find the equation of the line which passes through two points. The one I like the most is the vectorial method.

Let $v$ be the vector that connects two points ${P}_{0} = \left({x}_{0} , {y}_{0}\right)$ and ${P}_{1} = \left({x}_{1} , {y}_{1}\right)$ and $k$ a scalar, we can find any generic point $P = \left(x , y\right)$ in their line by adding $k \cdot v$ to the coordinates of ${P}_{0}$.

$P = {P}_{0} + k \cdot v \to \left(x , y\right) = \left({x}_{0} , {y}_{0}\right) + k \cdot v$
$v = \left({x}_{1} - {x}_{0} , {y}_{1} - {y}_{0}\right)$
$x = {x}_{0} + k \cdot \left({x}_{1} - {x}_{0}\right) \to k = \frac{\left(x - {x}_{0}\right)}{\left({x}_{1} - {x}_{0}\right)}$
$y = {y}_{0} + k \cdot \left({y}_{1} - {y}_{0}\right) \to k = \frac{\left(y - {y}_{0}\right)}{\left({y}_{1} - {y}_{0}\right)}$
$\frac{\left(x - {x}_{0}\right)}{\left({x}_{1} - {x}_{0}\right)} = \frac{\left(y - {y}_{0}\right)}{\left({y}_{1} - {y}_{0}\right)}$
$y = \left(x - {x}_{0}\right) \cdot \frac{\left({y}_{1} - {y}_{0}\right)}{\left({x}_{1} - {x}_{0}\right)}$//

Hope it helps.

Or

1) Find the slope using ...

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{1 - 3}{8 - \left(- 2\right)} = \frac{- 2}{8 + 2} = \frac{- 2}{10} = - \frac{1}{5}$

2) Find the $y$-intercept, $b$, by using the slope intercept equation, $y = m x + b$

Use either of the points. I will use the $\left(8 , 1\right)$.

$1 = \left(- \frac{1}{5}\right) \left(8\right) + b$

$1 = \left(- \frac{8}{5}\right) + b$

$1 + \left(\frac{8}{5}\right) = b$

$\frac{5}{5} + \left(\frac{8}{5}\right) = b$

$\frac{13}{5} = b$

3) Rewrite the slope intercept form, $y = m x + b$, and substitute in the $y$-intercept, $b$, and the slope, $m$.

$y = - \frac{1}{5} x + \frac{13}{5}$

In the image below see the function graphed. The green squares are the points (8,1) and (-2,3). Notice that they are on the line.