First we need to determine the slope of the line. The slope can be found by using the formula: #m = (color(red)(y_2) - color(blue)(y_1))/(color(red)(x_2) - color(blue)(x_1))#

Where #m# is the slope and (#color(blue)(x_1, y_1)#) and (#color(red)(x_2, y_2)#) are the two points on the line.

Substituting the values from the points in the problem gives:

#m = (color(red)(2) - color(blue)(8))/(color(red)(2) - color(blue)(5)) = (-6)/-3 = 2#

We can now use the point slope formula to write an equation for the line. The point-slope form of a linear equation is: #(y - color(blue)(y_1)) = color(red)(m)(x - color(blue)(x_1))#

Where #(color(blue)(x_1), color(blue)(y_1))# is a point on the line and #color(red)(m)# is the slope.

Substituting the slope we calculated and the values from either point in the problem (I will use the values from the second point) into the formula gives:

#(y - color(blue)(2)) = color(red)(2)(x - color(blue)(2))#

We can now solve this equation for the Standard Form of a Linear Equation. The standard form of a linear equation is: #color(red)(A)x + color(blue)(B)y = color(green)(C)#

Where, if at all possible, #color(red)(A)#, #color(blue)(B)#, and #color(green)(C)#are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

#y - color(blue)(2) = (color(red)(2) xx x) - (color(red)(2) xx color(blue)(2))#

#y - color(blue)(2) = 2x - 4#

#-color(red)(2x) + y - color(blue)(2) + 2 = -color(red)(2x) + 2x - 4 + 2#

#-2x + y - 0 = 0 - 2#

#-2x + y = -2#

#color(red)(-1)(-2x + y) = color(red)(-1) xx -2#

#(color(red)(-1) xx -2x) + (color(red)(-1) xx y) = 2#

#2x + (-1y) = 2#

#color(red)(2)x - color(blue)(1)y = color(green)(2)#