Geometry with Trigonometry

Sec. 6.4 Parametric equations of a line 89

6.4 Parametric equations of a line

6.4.1

Let l be a line with Cartesian equation ax+by+c=0.

(i)If Z 0 ≡(x 0 ,y 0 )is in l, then

l={Z≡(x,y):x=x 0 +bt,y=y 0 −at,(t∈R)}.

(ii)If Z 1 ≡(x 1 ,y 1 )=(x 0 +b,y 0 −a)and≤lis the natural order on l for which

Z 0 ≤lZ 1 , then for Z 2 ≡(x 0 +bt 2 ,y 0 −at 2 ),Z 3 ≡(x 0 +bt 3 ,y 0 −at 3 )we have

t 2 ≤t 3 if and only if Z 2 ≤lZ 3.

(iii)If Z 1 ≡(x 1 ,y 1 )=(x 0 +b,y 0 −a),then

[Z 0 ,Z 1 ]={Z≡(x,y):x=x 0 +bt,y=y 0 −at,( 0 ≤t≤ 1 )}.

(iv)With Z 1 as in(ii),

[Z 0 ,Z 1 ={Z≡(x,y):x=x 0 +bt,y=y 0 −at,(t≥ 0 )}.

Proof.
(i) IfZ∈lthenax+by+c= 0 ,ax 0 +by 0 +c=0, so that
b(y−y 0 )=−a(x−x 0 ). (6.4.1)
Whenb=0, let us definetbyt=(x−x 0 )/b; then by (6.4.1) we must have,y−y 0 =
−at. Thus
x=x 0 +bt,y=y 0 −at, (6.4.2)
for somet∈R.
Whenb=0thena=0, and by (6.4.1) we must havex=x 0 .Ifwedefinetby
t=(y−y 0 )/(−a), then we have (6.4.2) for somet∈R.
Conversely suppose that (6.4.2) holds for anyt∈R.Then
ax+by+c=a(x 0 +bt)+b(y 0 −at)+c=ax 0 +by 0 +c= 0.

(ii) We first suppose thatlis not perpendicular tom=OI,sothatb=0. We recall

thatZ 0 ,Z 1 are distinct points onlfor whichZ 0 ≤lZ 1 .Let≤mbe the natural order

onmfor whichO≤mI.LetU 0 =πm(Z 0 ),U 1 =πm(Z 1 )so thatU 0 ≡(x 0 , 0 ),U 1 ≡

(x 0 +b, 0 ).

U 0 U 1

Z 0 Z^1

Figure 6.4. Direct correspondence.

U 1 U 0

Z 0

Z 1

Indirect correspondence.