For this case we have that by definition, the equation of the line in the slope-intersection form is given by:
[tex]y = mx + b[/tex]
Where:
m: It is the slope of the line
b: It is the cut-off point with the y axis
We have the following points through which the line passes:
[tex](x_ {1}, y_ {1}): (- 6,7)\\(x_ {2}, y_ {2}): (- 3,6)[/tex]
We find the slope of the line:
[tex]m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {6-7} {- 3 - (- 6)} = \frac {-1} {-3 + 6} = \frac {-1} {3} = -\frac {1} {3}[/tex]
Thus, the equation of the line is of the form:
[tex]y = - \frac {1} {3} x + b[/tex]
We substitute one of the points and find b:
[tex]6 = - \frac {1} {3} (- 3) + b\\6 = 1 + b\\b = 5[/tex]
Finally, the equation is:
[tex]y = - \frac {1} {3} x + 5[/tex]
Answer:
[tex]y = - \frac {1} {3} x + 5[/tex]
