For this case we have that by definition, the equation of a line in the slope-intersection form is given by:
[tex]y = mx + b[/tex]
Where:
m: It's the slope
b: It is the cut point with the "y" axis
[tex]m = \frac {y2-y1} {x2-x1}[/tex]
Then, we have the points:
[tex](x1, y1): (- 6, -3)\\(x2, y2) :( 6, -7)[/tex]
Substituting:
[tex]m = \frac {-7 - (- 3)} {6 - (- 6)}\\m = \frac {-7 + 3} {6 + 6}\\m = \frac {-4} {12}\\m = - \frac {1} {3}[/tex]
Thus, the equation is:
[tex]y = - \frac {1} {3} x + b[/tex]
We substitute a point to find the cut point:
[tex]-7 = - \frac {1} {3} (6) + b\\-7 = -2 + b\\b = -7 + 2\\b = -5[/tex]
Finally the equation is:
[tex]y = - \frac {1} {3} x-5[/tex]
ANswer:
[tex]y = - \frac {1} {3} x-5[/tex]
