How do you find equations of the lines passing through (1,3 ) with slope -2/3?

Jul 1, 2016

slope-intercept form $y = - \frac{2}{3} x + \frac{11}{3}$

standard form $3 y - 2 x - 11 = 0$

Explanation:

To find the equation of the line passing through $\left(1 , 3\right) w i t h a s l o p e o f$-2/3#

We will use the point-slope formula $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

Where $m$ is the slope and $\left({x}_{1} , {Y}_{1}\right)$are the coordinates of the point on the line.

$m = - \frac{2}{3}$
${x}_{1} = 1$
${y}_{1} = 3$

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

$y - 3 = - \frac{2}{3} x + \frac{2}{3}$

$y \cancel{- 3} \cancel{+ 3} = - \frac{2}{3} x + \frac{2}{3} + 3$

$y = - \frac{2}{3} x + \frac{2}{3} + \frac{9}{3}$

$y = - \frac{2}{3} x + \frac{11}{3}$

$\left(3\right) y = - \left(\frac{2}{\cancel{3}} x + \frac{11}{\cancel{3}}\right) \cancel{3}$

$3 y = 2 x + 11$

$3 y - 2 x - 11 = \cancel{2 x} \cancel{+ 11} \cancel{- 2 x} \cancel{- 11}$

$3 y - 2 x - 11 = 0$