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 = (x1, y1),   and let   B = (x2, y2)   indicate the coordinates of   A   and   B.

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

ax1 + by1 = c

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

C = (1 - t)A + tB

= (1 - t)(x1, y1) + t(x2, y2)

= ((1 - t)x1 + tx2, (1 - t)y1 + ty2)

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.

a[(1 - t)x1 + tx2] + b[(1 - t)y1 + ty2]

Remove parentheses.

= ax1 - atx1 + atx2 + by1 - bty1 + bty2

Rearrange the terms.

= ax1 + by1 + atx2 + bty2 - atx1 - bty1

Factor.

= ax1 + by1 + t(ax2 + by2 - (ax1 + by1))

Substitute   ax1 + by1 = c

= c + t(ax2 + by2 - c)

If   ax2 + by2 > c,   then

ax2 + by2 - c > 0

and since   t > 0,

t(ax2 + by2 - c) > 0

so

c + t(ax2 + by2 - c) > c,

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

On the other hand, if   ax2 + by2 > c,   then

ax2 + by2 - c < 0

and since   t > 0,

t(ax2 + by2 - c) < 0

so

c + t(ax2 + by2 - 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.

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