# What is the equation of the line with slope  m= 19/25  that passes through  (16/5 73/10) ?

Aug 17, 2017

$y - \frac{73}{10} = \frac{19}{25} \left(x - \frac{16}{5}\right) \leftarrow$ Point-slope form

$y = \frac{19}{25} x + \frac{1217}{250} \leftarrow$ y=mx+b form

$- \frac{19}{25} x + y = \frac{1217}{250} \leftarrow$ Standard form

#### Explanation:

Seeing how we already have the slope and a coordinate, we can find the equation of the line by using the point-slope formula: $y - {y}_{1} = m \left(x - {x}_{1}\right)$ where $m$ is the slope $\left(m = \frac{19}{25}\right)$ and $\left({x}_{1} , {y}_{1}\right)$ is a point on the line. Thus, $\left(\frac{16}{5} , \frac{73}{10}\right) \to \left({x}_{1} , {y}_{1}\right)$.

The equation is then...

$y - \frac{73}{10} = \frac{19}{25} \left(x - \frac{16}{5}\right)$

...in point slope form.

Since you did not specify in what form the equation should be expressed, the above is an acceptable answer but we could also rewrite the equation is $y = m x + b$ form. To do this, we solve for $y$.

$y - \frac{73}{10} = \frac{19}{25} x - \frac{304}{125}$

$y \cancel{- \frac{73}{10} + \frac{73}{10}} = \frac{19}{25} x - \frac{304}{125} + \frac{73}{10}$

$y = \frac{19}{25} x - \left[\frac{304}{125} \left(\frac{2}{2}\right)\right] + \left[\frac{73}{10} \left(\frac{25}{25}\right)\right]$

$y = \frac{19}{25} x - \frac{608}{250} + \frac{1825}{250}$

$y = \frac{19}{25} x + \frac{1217}{250} \leftarrow$ The equation in y=mx+b form

Alternatively, the equation could also be expressed in standard form: $A x + B y = C$

$- \frac{19}{25} x + y = \cancel{\frac{19}{25} x - \frac{19}{25} x} + \frac{1217}{250}$

$- \frac{19}{25} x + y = \frac{1217}{250} \leftarrow$ The equation is standard form