Theorem 2.12: Let   A,   B   and   C   be three noncolinear points, and let

D = qA + rB + sC

where

q + r + s = 1

be a point in the plane. Then   D   is on the same side of   BC   as   A   if and only if   q   is positive.

Proof: Let

A = (x0, y0)

B = (x1, y1)

C = (x2, y2)

and

D = (x, y)

Then

D = qA + rB + sC

= q(x0, y0) + r(x1, y1) + s(x2, y2)

= (qx0 + rx1 + sx2, qy0 + ry1 + sy2)

Finally let

ax + by = c

be the equation for the line determined by   BC.   If we check the coordinates for the point   D   in the left side of this equation we get

a(qx0 + rx1 + sx2) + b(qy0 + ry1 + sy2)

Remove parentheses

= aqx0 + arx1 + asx2 + bqy0 + bry1 + bsy2

These terms can be rearranged as

= q(ax0 + by0) + r(ax1 + by1) + s(ax2 + by2)

Since  B  and   C   are on the line,

ax1 + by1 = c

and

ax2 + by2 = c

If we substitute these into the last equation, we get

= q(ax0 + by0) + rc + sc

or

= q(ax0 + by0) + (r + s)c

However, since   q + r + s = 1,

r + s = 1 - q

Our equation then becomes

 

= q(ax0 + by0) + (1 - q)c

or

= q(ax0 + by0 - c) + c

This will satisfy the same inequality as the point   A   if and only if   q   is positive.

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next theorem (2.13)