For this case we have that the equation of a straight line in the form of an intersection is given by:
[tex]y = mx + b[/tex]
Where:
m: It's the slope
b: It is the cutoff point with the y axis
We found the slope:
[tex]m = \frac {y2-y1} {x2-x1} = \frac {8-6} {4-5} = \frac {2} {- 1} = - 2[/tex]
So, the line is:
[tex]y = -2x + b[/tex]
We find the cut point by substituting a point:
[tex]8 = -2 (4) + b\\8 = -8 + b\\b = 8 + 8\\b = 16[/tex]
Finally, the equation is:
[tex]y = -2x + 16[/tex]
We can also have the equation of the point-slope form:
[tex]y-y_ {0} = m (x-x_ {0})[/tex]
Where:
[tex](x_ {0}, y_ {0}) = (4,8)[/tex]represents a point:
So:
[tex]y-8 = -2 (x-4)[/tex]
ANswer:
[tex]y-8 = -2 (x-4)\\y = -2x + 16[/tex]