For this case we have that by definition, the equation of the line of the slope-intersection form is given by:
[tex]y = mx + b[/tex]
Where:
m: It's the slope
b: It is the cut-off point with the y axis
According to the graph, we place two points through which the line passes:
[tex](x_ {1}, y_ {1}) :( 6,0)\\(x_ {2}, y_ {2}) :( 0, -3)[/tex]
We found the slope:
[tex]m = \frac{y_ {2} -y_ {1}} {x_ {2} -x_ {1}}[/tex]
Substituting we have:
[tex]m = \frac {-3-0} {0-6} = \frac {-3} {- 6} = \frac {1} {2}[/tex]
Thus, the equation is of the form:
[tex]y = \frac {1} {2} x + b[/tex]
We substitute one of the points and find the cut-off point:
[tex]0 = \frac {1} {2} (6) + b\\0 = 3 + b\\-3 = b\\b = -3[/tex]
Finally, the equation is:
[tex]y = \frac {1} {2} x-3[/tex]
ANswer:
[tex]y = \frac {1} {2} x-3[/tex]
