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)(2))/(color(red)(5) - color(blue)(3)) > 4/2 = 2#

We can next 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 first point in the problem gives:

#(y - color(red)(2)) = color(blue)(2)(x - color(red)(3))#

We can now transform this to the Standard Form for 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(red)(2) = (color(blue)(2) xx x) - (color(blue)(2) xx color(red)(3))#

#y - color(red)(2) = 2x - 6#

First, we add #color(blue)(2)# and subtract #color(red)(2x)# from each side of the equation to isolate the #x# and #y# variables on the left side of the equation and the constant on the right side of the equation while keeping the equation balanced:

#-color(red)(2x) + y - color(red)(2) + color(blue)(2) = -color(red)(2x) + 2x - 6 + color(blue)(2)#

#-2x + y - 0 = 0 - 4#

#-2x + y = -4#

Now, we multiply each side of the equation by #color(red)(-1)# to ensure the coefficient for the #x# variable is positive while keeping the equation balanced:

#color(red)(-1)(-2x + y) = color(red)(-1) xx -4#

#(color(red)(-1) xx -2x) + (color(red)(-1) xx y) = 4#

#color(red)(2)x - color(blue)(1)y = color(green)(4)#