First, we must 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)(15) - color(blue)(9))/(color(red)(5) - color(blue)(3)) = 6/2 = 3#

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 first point from the problem gives:

#(y - color(red)(9)) = color(blue)(3)(x - color(red)(3))#

We can also substitute the slope we calculated and the second point from the problem giving:

#(y - color(red)(15)) = color(blue)(3)(x - color(red)(5))#

The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value. We can take the last equation and solve for #y#:

#y - color(red)(15) = (color(blue)(3) xx x) - (color(blue)(3) xx color(red)(5))#

#y - color(red)(15) = 3x - 15#

#y - color(red)(15) + 15 = 3x - 15 + 15#

#y = color(red)(3)x + color(blue)(0)#

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

Transforming the last equation gives:

#-3x + y = -3x + color(red)(3)x + color(blue)(0)#

#-3x + y = 0 + color(blue)(0)#

#-3x + y = 0#

#color(red)(-1)(-3x + y) = color(red)(-1) xx 0#

#color(red)(3)x + color(blue)(-1)y = color(green)(0)#