Question

How do you write an equation in standard form for a line passing through R(3, 3), S(-6, -6)?

Answer 1

Method 1.
Notice that the values of X-coordinate and Y-coordinate are equal in two different points.
So, an intelligent guess would be to consider an equation
`y=x`
as the representation of this line.

Method 2.
Assume that the equation we are looking for is
`y=a*x+b`
where `a` and `b` are unknown coefficients that we have to determine using the data provided.
Substitute the given coordinates into this equation:
`3=a*3+b`
`-6=a*(-6)+b`

Resolve the first equation for `b` in terms of `a` and substitute into the second equation to have only one equation with one unknown `a`:
`b=3-3a`
`-6=-6a+(3-3a)`

From the latter we derive:
`-9=-6a-3a`
`-9=-9a`
`a=1`

Now find `b` that we expressed in terms of `a`:
`b=3-3a=3-3*1=0`

The above resulted in the equation for our line:
`y=1*x+0` or, simply, `y=x`.

Method 3.
Again, looking for an equation in a format
`y=a*x+b`
As it's known, the coefficient `a` is a slope of a line that can be calculated as a ratio between increment of ordinate (Y-coordinate changes from `-6` to `3`) to increment of abscissa (X-coordinate changes from `-6` to `3`)).
In our case this ratio equal to:
`a=(3-(-6))/(3-(-6))=1`

To determine coefficient `b`, we can substitute the values of abscissa and ordinate of one of the points to get an equation for `b`:
`3 = 1*3+b`
from which follows that `b=0`.
So, our equation is
`y=1*x+0` or, simply, `y=x`.