# How do you write an equation of a line with points (5/12, -1), (-3/4, 1/6)?

Mar 12, 2017

12x+12y+7 = 0

#### Explanation:

We know equation of a line = $\left[\frac{y - {y}_{1}}{x - {x}_{1}}\right] = \left[\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}\right]$

here ${x}_{1} , {y}_{1} = \frac{5}{12} , - 1 \mathmr{and} {x}_{2} , {y}_{2} = - \frac{3}{4} , \frac{1}{6}$

hence, $\left[\frac{y - \left(- 1\right)}{x - \frac{5}{12}}\right] = \left[\frac{\frac{1}{6} - \left(- 1\right)}{- \frac{3}{4} - \frac{5}{12}}\right]$

$\Rightarrow \left[\frac{y + 1}{x - \frac{5}{12}}\right] = \left[\frac{\frac{1}{6} + 1}{\frac{- 9 - 5}{12}}\right]$

$\Rightarrow \left[\frac{y + 1}{x - \frac{5}{12}}\right] = \frac{\frac{7}{6}}{- \frac{14}{12}}$

$\Rightarrow \left[\frac{y + 1}{x - \frac{5}{12}}\right] = \frac{7}{6} \times - \frac{12}{14}$

$\Rightarrow \left[\frac{y + 1}{x - \frac{5}{12}}\right] = - 1$

$\Rightarrow y + 1 = - \left(x - \frac{5}{12}\right)$

$\Rightarrow y + 1 + x - \frac{5}{12} = 0$

$\Rightarrow x + y + \frac{7}{12} = 0$

$\Rightarrow 12 x + 12 y + 7 = 0$