# What is the equation of the line that passes through (2, 1) and (5, -1)?

Jul 6, 2016

$y = \frac{- 2}{3} x + \frac{7}{3}$

#### Explanation:

Since we have two points the first thing I would do is calculate the gradient of the line.

We can use the formula gradient(m) $= \frac{\Delta y}{\Delta x} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

We then need to select our values to substitute into the equation, for this we will take our first point $\left(2 , 1\right)$ and make ${x}_{1} = 2$ and ${y}_{1} = 1$. Now take the second point $\left(5 - 1\right)$ and make ${x}_{2} = 5$ and ${y}_{2} = - 1$. Simply substitute the values in the equation:

gradient(m) $= \frac{\Delta y}{\Delta x} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{- 1 - 1}{5 - 2} = \frac{- 2}{3}$

Now that we have the gradient substitute that into $y = m x + c$ so that $y = \frac{- 2}{3} x + c$

To find $c$ we need to use one of the given points, so substitute one of these points into our equation: $y = \frac{- 2}{3} x + c$ In this explanation we will use $\left(2 , 1\right)$. So $1 = \frac{- 2}{3} \left(2\right) + c$

Now solve as a linear equation to obtain $c$:

$1 = \frac{- 4}{3} + c$
$1 - \frac{- 4}{3} = c$
$\frac{7}{3} = c$
$c = \frac{7}{3}$

Substitute the value for $c$ into the equation: $y = \frac{- 2}{3} x + c$ so that $y = \frac{- 2}{3} x + \frac{7}{3}$