How do you write the equation of the line through (4,-8) and (8,5)?

Step 1: Determine the slope of the line through the given points

#"slope"=("change in "y)/("change in "x)#

Given the points, #(x_1,y_1)=(4,-8)# and #(x_2,y_2)=(8,5)#,
the slope, #m# is given by:
#color(white)("XXX")m=(5-(-8))/(8-4)=13/4#

Step 2: Write the equation in slope-point form
Given a slope #m# and a point #(x_1,y_1)#,
the slope-point form of the equation is:
#color(white)("XXX")y-y_1=m(x-x_1)#
In this case we have:
#color(white)("XXX")m=13/4" and "(x_1,y_1)=(4,-8)#
giving
#color(white)("XXX")y-(-8)=13/4(x-4)#

Step 3: Convert into standard form
Standard form of a linear equation is
#color(white)("XXX")Ax+By=C#
Starting from the slope-point form (above) we have:
#color(white)("XXX")4(y+8)=13(x-4)#

#color(white)("XXX")4y+32=13x-52#

#color(white)("XXX")13x-4y=84#

For verification purposes, here is a graph with the given points and the equation #13x-4y=84#

#color(blue)("Preamble")#

consider the standardised equation format of: #y=mx+c#
where #m# is the gradient.

#m=("changing in up or down")/("change in along")#

Very important#->#the change in along (usually the x-axis) is read left to right.

Watch out for this because sometimes questions give the points in the revers order.

So the first point is at say #x_1# and the second point is at say #x_2#. Then #x_1 < x_2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

#color(brown)("Determine the gradient")#

Let point 1 be #P_1->(x_1,y_1)=(4,-8)#
Let point 2 be #P_2->(x_2,y_2)=(8,5)#

#m=(y_2-y_1)/(x_2-x_1) = (5-(-8))/(8-4) =(5+8)/(8-4)=13/4#

So we now have:#" "y=13/4x+c#

#color(brown)("Determine the value of the constant "c)#

I chose #P_1 ->#using the value for this point substitute for #x and y#

#y=13/4x+c" "->" "-8=13/4(4)+c #

#c=-8-13=-21#

So we now have:#" "y=13/4x-21#