For this case we have that by definition, the equation of a line in standard form is given by:
[tex]ax + by = c[/tex]
On the other hand, we have that 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-off point with the y axis
We have two points:
[tex](x_ {1}, y_ {1}): (\frac {1} {2}, 2)\\(x_ {2}, y_ {2}): (-3, -1)[/tex]
We find the slope of the line:
[tex]m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-1-2} {- 3- \frac {1} {2}} = \frac {-3} {- \frac {7} {2}} = \frac {6} {7}[/tex]
Thus, the equation is of the form:
[tex]y = \frac {6} {7} x + b[/tex]
We substitute one of the points and find "b":
[tex]-1 = \frac {6} {7} (- 3) + b\\-1 = - \frac {18} {7} + b\\-1+ \frac {18} {7} = b\\b = \frac {-7 + 18} {7}\\b = \frac {11} {7}[/tex]
Finally, the equation is:
[tex]y = \frac {6} {7} x + \frac {11} {7}[/tex]
We convert the equation to the standard form:
[tex]y- \frac {11} {7} = \frac {6} {7} x\\7 (y- \frac {11} {7}) = 6x\\7y-11 = 6x\\-6x + 7y = 11\\6x-7y = -11[/tex]
ANswer:
[tex]6x-7y = -11[/tex]
