How do you write the equation of a line in point slope form and slope intercept form given points (-1, -5) (-7, -6)?

Mar 20, 2018

The equation of the line in slope intercept form is $y = \frac{1}{6} x - \frac{29}{6.}$
The equation of line in point slope form is $y + 5 = \frac{1}{6} \left(x + 1\right)$

Explanation:

The slope of the line passing through

$\left(- 1 , - 5\right) \mathmr{and} \left(- 7 , - 6\right)$ is

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{- 6 + 5}{- 7 + 1} = \frac{- 1}{-} 6 = \frac{1}{6}$

Let the equation of the line in slope intercept form be

$y = m x + c \mathmr{and} y = \frac{1}{6} \cdot x + c$. The point $\left(- 1 , - 5\right)$ will satisfy

the equation .So, $- 5 = \frac{1}{6} \cdot \left(- 1\right) + c$ or

$c = - 5 + \frac{1}{6} = \frac{- 29}{6}$. Hence the equation of the line in

slope intercept form is $y = \frac{1}{6} x - \frac{29}{6.}$

The equation of line in point slope form passing through

$\left({x}_{1} , {y}_{1}\right)$ having slope of $m$ is $y - {y}_{1} = m \left(x - {x}_{1}\right)$.

The equation of line passing through $\left(- 1 , - 5\right)$ having slope of

$\frac{1}{6}$ is $y + 5 = \frac{1}{6} \left(x + 1\right)$ [Ans]