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 statement we have to:
[tex]m = \frac {1} {2}[/tex]
So, the equation is of the form:
[tex]y = \frac {1} {2} x + b[/tex]
We substitute the given point and find "b":
[tex]-7 = \frac {1} {2} (2) + b\\-7 = 1 + b\\-7-1 = b\\b = -8[/tex]
Finally, the equation is:
[tex]y = \frac {1} {2} x-8[/tex]
By definition, the standard form of the equation of the line is:
[tex]ax + by = c[/tex]
Then, we manipulate the equation algebraically:
[tex]y = \frac {1} {2} x-8\\y + 8 = \frac {1} {2} x\\2 (y + 8) = x\\2y + 16 = x\\x-2y = 16[/tex]
Answer:
The equation in the standard form is:
[tex]x-2y = 16[/tex]
