**Theorem
2.5**: Let * A* be a point on the
line determined by the
equation *ax* + *by* = *c*, and let *B *be a point not on that line.
Then the points on the
line determined by * A* and *B*,
which are on the same side
of *A* as *B*, are on the same
side of the line *ax* +
*by* = *c* as *B*, and the points on the
other side of *A* from *B* on
the line determined by *A* and *B* are on the other side
of the line *ax* + *by* = *c*.

**Proof**: Let * A* = (*x*_{1}, *y*_{1}), and let *B* =
(*x*_{2}, *y*_{2}) indicate the coordinates of *A* and *B*.

Since *A* is on the line
*ax* + *by* = *c*, we have

Let * C* be any point on
the line determined by *A* and *B*. Then by Theorem 2.1, there is a real
number *t* such that

If we want to see on which
side of the line * ax* + *by* = *c* this point is, we substitute its coordinates into the left side of
the equation determining the line.

Remove parentheses.

Rearrange the terms.

Factor.

Substitute * ax*_{1} + *by*_{1} = *c*

If *ax*_{2} + *by*_{2} > *c*, then

*ax*_{2} + *by*_{2} - *c* > 0

and since *t* > 0,

*t*(*ax*_{2} + *by*_{2} - *c*) > 0

so

*c* + *t*(*ax*_{2} + *by*_{2} - *c*) > *c*,

and *C* is on the same side of *A* as *B*.

On the other hand, if *ax*_{2} + *by*_{2} > *c*, then

*ax*_{2} + *by*_{2} - *c* < 0

and since *t* > 0,

*t*(*ax*_{2} + *by*_{2} - *c*) < 0

so

*c* + *t*(*ax*_{2} + *by*_{2} - *c*) < *c*,

and *C* is on the same side of *A* as *B*.

Iff * t* < 0, then these inequalities will be reversed and *C* is on the other side of *A* from *B*.