Theorem 2.1: (The parametric representation of a line) Given two points   (x1, y1)   and   (x2, y2),   the point   (x, y)   is on the line determined by   (x1, y1)   and   (x2, y2)   if and only if there is a real number   t   such that

x = (1 - t)x1 + tx2,

and

y = (1 - t)y1 + ty2

 

Proof: Assume that there is a real number   t   such that

x = (1 - t)x1 + tx2,

and

y = (1 - t)y1 + ty2

Then

y - y1 = (1 - t)y1 + t y2 - y1

Remove parentheses

= y1 - ty1 + ty2 - y1

= -ty1 + ty2

= t(y2 - y1)

and

x - x1 = (1 - t)x1 + tx2 - x1

= x1 - tx1 + tx2 - x1

= - tx1 + tx2

= t(x2 - x1)

If the points are vertical, then    x2 - x1 = 0   so

x - x1 = t(x2 - x1) = 0

so   x = x1 = x2,  and the point is on the same vertical line.

Otherwise, let   m   be the slope of the line between   (x1, y1)   and   (x2, y2).   Then

and

y - y1 = m(x - x1)

and   (x, y)   satisfies the point-slope form of the equation of the line going through   (x1, y1)   with slope   m, by Theorem 1.3.

Conversely, assume that   (x, y)   is any point on the line. We will first dispose of the case where the line is vertical. In this case,

x = x1 = x2

Let

We can assume the denominator is nonzero because we have two distinct points.

Then

t(y2 - y1) = y - y1

so

(1 - t)y1 + ty2

= y1 + t(y2 - y1)

= y1 + y - y1

= y

and

(1 - t)x1 + tx2

= (1 - t)x + tx

= x

We must also dispose of the case of a horizontal line where   y2 = y1.   Let

We can again assume the denominator is nonzero because we have two distinct points.

Then

t(x2 - x1) = x - x1

so

(1 - t)x1 + tx2

= x1 + t(x2 - x1)

= x1 + x - x1

= x

and

(1 - t)y1 + ty2

= (1 - t)y + ty

= y

Otherwise, the slope of the line is given by

by Theorem 1.1. Multiply both sides by   x - x1   and divide by    y2 - y1

We can assume that the denominators on both sides are nonzero because we have disposed of those cases.

Define

Then

x - x1 = t(x2 - x1)

and

y - y1 = t(y2 - y1)

So

x = x1 + t(x2 - x1)

and

y = y1 + t(y2 - y1)

or

x = (1 - t)x1 + tx2

and

y = (1 - t)y1 + ty2

 

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