# How do you write the standard form of a line given (12, 1) and has a slope of 4/3?

Sep 15, 2016

$4 x - 3 y - 45 = 0$

#### Explanation:

The equation of a line given a point $\left({x}_{1} , {y}_{1}\right)$ and slope $m$ is given by

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

Hence, equation of a line given a point $\left(12 , 1\right)$ and slope $\frac{4}{3}$ is

$\left(y - 1\right) = \frac{4}{3} \times \left(x - 12\right)$ or

$3 \left(y - 1\right) = 4 \left(x - 12\right)$ or

$3 y - 3 = 4 x - 48$ or

$4 x - 3 y - 45 = 0$, which is in standard form $a x + b y + c = 0$