First, we can obtain an equation for the line using the point-slope formula. 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 values from the problem gives:

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

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

We can begin solving for this form by multiplying each side of the equation by #color(purple)(3)# to eliminate the fraction:

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

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

#3y - 15 = 5x - 15#

Next we can add #color(red)(15)# and subtract #color(blue)(5x)# to each side of the equation to isolate the constant on the right side of the equation while maintaining the balance of the equation.

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

#-5x + 3y - 0 = 0 - 0#

#-5x + 3y = 0#

Now, we can multiple each side of the equation by #color(red)(-1)# to make the coefficient of #x# positive or non-negative.

#color(red)(-1)(-5x + 3y) = color(red)(-1) xx 0#

#5x - 3y = 0#