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`