# How do you write an equation of a line with point (5/4, 1), (-1/4, 3/4)?

Jun 9, 2017

$y = \frac{1}{6} + \frac{19}{24}$

#### Explanation:

Given -

$\left({x}_{1} , {y}_{1}\right) = \left(\frac{5}{4} , 1\right)$
$\left({x}_{2} , {y}_{2}\right) = \left(- \frac{1}{4} , \frac{3}{4}\right)$

$y - {y}_{1} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \left(x - {x}_{1}\right)$

$y - 1 = \frac{\left(\frac{3}{4}\right) - 1}{\left(- \frac{1}{4}\right) - \frac{5}{4}} \left(x - \frac{5}{4}\right)$

$y - 1 = \frac{\frac{3 - 4}{4}}{\frac{- 1 - 5}{4}} \left(x - \frac{5}{4}\right)$

$y - 1 = \frac{\frac{- 1}{4}}{\frac{- 6}{4}} \left(x - \frac{5}{4}\right)$
$y - 1 = \frac{\frac{1}{4}}{\frac{6}{4}} \left(x - \frac{5}{4}\right)$
$y - 1 = \frac{1}{\cancel{4}} \times \frac{\cancel{4}}{6} \left(x - \frac{5}{4}\right)$
$y - 1 = \frac{1}{6} \left(x - \frac{5}{4}\right)$
$y - 1 = \frac{1}{6} x - \frac{5}{24}$
$y = \frac{1}{6} x - \frac{5}{24} + 1$
$y = \frac{1}{6} x + \left(\frac{- 5 + 24}{24}\right)$
$y = \frac{1}{6} x + \left(\frac{19}{24}\right)$
$y = \frac{1}{6} + \frac{19}{24}$