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)(6) - color(blue)(4))/(color(red)(0) - color(blue)(-2)) = (color(red)(6) - color(blue)(4))/(color(red)(0) + color(blue)(2)) = 2/2 = 1#

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

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

Substituting the slope we calculated and the values from the second point in the problem gives:

#(y - color(red)(6)) = color(blue)(1)(x - color(red)(0))#

#y - color(red)(6) = 1x#

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

To start transforming the equation we wrote to standard form we can add #color(red)(6)# and subtract #color(blue)(1x)# from each side of the equation to put both variables on the left side of the equation and the constant on the right side of the equation:

#-color(blue)(1x) + y - color(red)(6) + color(red)(6) = - color(blue)(1x) + 1x + color(red)(6)#

#-1x + y - 0 = 0 + 6#

#-1x + y = 6#

Now, because the coefficient of the #x# variable should be a positive integer we can multiply each side of the equation by #color(red)(-1)# to change the sign of the #x# variable while keeping the equation balanced:

#color(red)(-1)(-1x + y) = color(red)(-1) xx 6#

#(color(red)(-1) xx -1x) + (color(red)(-1) xx y) = -6#

#color(red)(1)x + color(blue)(-1)y = color(green)(-6)#