# How do you write the equation of a line in slope intercept, point slope and standard form given (1,-2), and (3,-8)?

Jan 17, 2018

#### Explanation:

The slope intercept form of equation of a line is $y = m x + c$, where $m$ is the slope of the line and $c$ is its intercept on $y$-axis.

Point slope form of equation of a line is $y - {y}_{1} = m \left(x - {x}_{1}\right)$, where $m$ is the slope of the line and $\left({x}_{1} , {y}_{1}\right)$ are the coordinates of the point through which the line passes.

The general form of equation of a line is $a x + b y + c = 0$.

For the equation of a line passing through two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$, the formula is $\frac{y - {y}_{1}}{{y}_{2} - {y}_{1}} = \frac{x - {x}_{1}}{{x}_{2} - {x}_{1}}$. However, we first find the slope of line $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$, which is $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$. For points $\left(1 , - 2\right)$ and $\left(3 , - 8\right)$ is $\frac{- 8 - \left(- 2\right)}{3 - 1} = - \frac{6}{2} = - 3$.

Hence selecting the point slope form of equation of desired line (choosing point $\left(1 , - 2\right)$) is $y - \left(- 2\right) = \left(- 3\right) \left(x - 1\right)$ or $y + 2 = - 3 x + 3$,

which in general form is $3 x + y - 1 = 0$

Note that had we chosen $\left(3 , - 8\right)$, the point slope form of equation could also have been $y - \left(- 8\right) = \left(- 3\right) \left(x - 3\right)$ or $y + 8 = - 3 x + 9$ which again gives us $3 x + y - 1 = 0$.

We can also write the equation as $y = - 3 x + 1$ in slope intercept form which shows slope as $- 3$ and $y$-intercept as $1$.

graph{-3x+1 [-10, 10, -5, 5]}