8. Write an equation in standard form for a linethat passes through (2, 2) and (12, -3).

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:[tex]y = mx + b[/tex]Where:m: It's the slopeb: It is the cut-off point with the y axisAccording to the statement we have two points through which the line passes:[tex](x_ {1}, y_ {1}): (2,2)\\(x_ {2}, y_ {2}): (12, -3)[/tex]We found the slope:[tex]m = \frac {y_ {2} -y- {1}} {x_ {2} -x_ {1}} = \frac {-3-2} {12-2} = \frac {-5} {10 } = - \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 "b":[tex]2 = - \frac {1} {2} (2) + b\\2 = -1 + b\\2 + 1 = b\\b = 3[/tex]Finally, the equation is of the form:[tex]y = - \frac {1} {2} x + 3[/tex]We write the equation of the standard form[tex]ax + by = c[/tex][tex]y-3 = -\frac {1} {2} x\\2y-6 = -x\\x + 2y = 6[/tex]ANswer:[tex]x + 2y = 6[/tex]