How do you write an equation in standard form for a line passing through (5,22) and (3,12)?

#"Write the equation in slope-intercept form"# #color(white)(........)##y = mx + b#

#color(white)(.........)#―――――――

First find #m#, the slope

Slope = #((y - y'))/((x - x'))#

1) Decide which point will be #(x,y)#

Assign #(5,22)# to be #(x,y)# Assign #(3,12)# to be #(x',y')#

You can choose either point to be #(x',y')# and it will come out the same, but if you select wisely, you can sometimes avoid working with negative numbers.

#((y - y'))/((x - x'))#

#((22 - 12))/((5 - 3))#

3) Do the subtractions to find the slope #m#

#color(white)(.....)##(10)/(2)# #=5# #larr# value for slope #m#

#color(white)(.....)##y = 5x + b#

#color(white)(.........)#―――――――

Next find #b#, the y intercept

1) Using either given point, sub in the values for #(x,y)# and solve for #b#

#color(white)(.......)## y = 5  x  + b# #color(white)(.....)##12 = 5(3) + b#

2) Clear the parentheses #12 = 15 + b#

3) Subtract #15# from both sides to isolate #b# #-3 = b#

So the equation in point-slope form is #color(white)(.....)##y = 5x - 3#

#color(white)(.........)#―――――――

#color(white)(.......)##ax + by = c# where #a# is a positive integer

#color(white)(.....)##y = 5x - 3#

1) Subtract #5x# from both sides to isolate the constant

#-5x + y = - 3#

2) Multiply through by #-1# to clear the minus sign on #x#

#5x - y = 3# #larr# standard form

#color(blue)("Preamble")#

Gradient (slope) is:

#m=(y-("known value for "y))/(x-("known value of "x)) ..Equation(1)#

Method
Step 1: determine #m# from the given points.
Step 2: use one of the given points to substitute values for #x and y# into #Equation(1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

#color(brown)("Step 1:")#

Always read left to right on the x-axis for this:

Set the left most point as #P_1->(x_1,y_1)=(3,12)#
Set the rightmost point as #P_2->(x_2,y_2)=(5,22)#

Set the gradient (slope) as #m#

#m=("change in the y-axis")/("change in the x-axis") ->(y_2-y_1)/(x_2-x_1)=(22-12)/(5-3)=10/2#

But #10/2# is the same as #5/1->5#

#m=5/1 larr" Deliberately written this way"#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Step 2:")#

Using #P_1#

#m=("change in the y-axis")/("change in the x-axis")->m=5/1=(y-12)/(x-3)#

Cross multiply giving:

#5(x-3)=1(y-12)#

#5x-15=y-12#

#5x-y=15-12#

#5x-y=3#