For this case we have that by definition, the equation of the line of the slope-intersection form is given by:
![y = mx + b](https://img.qammunity.org/2020/formulas/mathematics/high-school/fc4cgm6covys37zv2opmmp9ps4jxyjepvh.png)
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:
![(x_ {1}, y_ {1}) :( 6,0)\\(x_ {2}, y_ {2}) :( 0, -3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/w6pnd207vt4kzwzzpbtkar14i8wdest74a.png)
We found the slope:
![m = \frac{y_ {2} -y_ {1}} {x_ {2} -x_ {1}}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/r70cy6899eou03t9r25qngyp4r0e1cg533.png)
Substituting we have:
![m = \frac {-3-0} {0-6} = \frac {-3} {- 6} = \frac {1} {2}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/fbmplqyov4pkgnrxoscn3mcxpzphg9cgb8.png)
Thus, the equation is of the form:
![y = \frac {1} {2} x + b](https://img.qammunity.org/2020/formulas/mathematics/high-school/aps4k1ytzdl7lvyrq78c2xr52lpemskva6.png)
We substitute one of the points and find the cut-off point:
![0 = \frac {1} {2} (6) + b\\0 = 3 + b\\-3 = b\\b = -3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ylg6hgognb0cfm4w6aqw3lxi6p5zfxqowj.png)
Finally, the equation is:
![y = \frac {1} {2} x-3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/v0eqspda28qmu3ank3ekweqajb78q3tp8q.png)
ANswer:
![y = \frac {1} {2} x-3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/v0eqspda28qmu3ank3ekweqajb78q3tp8q.png)