What is the equation of the line between #(3,-13)# and #(-7,1)#?

When you know the coordinates of two points #P_1 = (x_1,y_1)# and #P_2 = (x_2,y_2)#, the line passing through them has equation

#\frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}#

Plug your values to get

#\frac{y+13}{1+13} = \frac{x-3}{-7-3} \iff \frac{y+13}{14} = \frac{x-3}{-10}#

Multiply both sides by #14#:

#y+13 = -\frac{7}{5}x + \frac{42}{10}#

Subtract #13# from both sides:

#y = -\frac{7}{5}x - 44/5#

Using the gradient (slope)

Reading left to right on the x-axis.
Set point 1 as #P_1->(x_1,y_1)=(-7,1)#
Set point 2 as #P_2->(x_2,y_2)=(3,-13)#

In reading this we 'travel' from #x_1# to #x_2# so to determine the difference we have #x_2-x_1 and y_2-y_1#

#color(red)(m)=("change in y")/("change in x") ->(y_2-y_1)/(x_2-x_1)=(-13-1)/(3-(-7)) = color(red)((-14)/(+10)=-7/5) #

We may choose any of the two: #P_1" or "P_2# for the next bit. I choose #P_1#

#m=-7/5=(y_2-1)/(x_2-(-7)) =(y_2-1)/(x_2+7)#

#-7(x_2+7)=5(y_2-1)#

#-7x_2-49=5y_2-5#

Add 5 to both sides

#-7x_2-44=5y_2#

Divide both sides by 5

#-7/5x_2-44/5=y_2#

Now using generic #x and y#

#-7/5x-44/5=y#

#y=-7/5x-44/5#