Geometry with Trigonometry

92 Cartesian coordinates; applications Ch. 6

(iv)

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

Proof. By 6.3.1, in the above we can takea=−(y 1 −y 0 ),b=x 1 −x 0 and the
conclusions follow immediately.

NOTE. We refer to

x=x 0 +bt,y=y 0 −at,(t∈R)

in 6.4.1 asparametric equationsof the linel,andtas theparameterof the point
Z≡(x,y).

6.5 Perpendicularity and parallelism of lines

6.5.1

Let l≡ax+by+c= 0 ,m≡a 1 x+b 1 y+c 1 = 0.

(i)Then l⊥m if and only if

aa 1 +bb 1 = 0. (6.5.1)

(ii)Also l‖m if and only if

ab 1 −a 1 b= 0. (6.5.2)

Proof.
(i) Suppose thatl⊥m.Thenlmeetsmin a unique point which we denote byZ 0.
By 6.4.1Z 1 ≡(x 0 +b,y 0 −a)is a point ofland similarlyZ 2 ≡(x 0 +b 1 ,y 0 −a 1 )is
a point ofm. Now by Pythagoras’ theorem|Z 0 ,Z 1 |^2 +|Z 0 ,Z 2 |^2 =|Z 1 ,Z 2 |^2 andsoby
6.1.1
[b^2 +(−a)^2 ]+[b^21 +(−a 1 )^2 ]=(b−b 1 )^2 +(a 1 −a)^2.

This simplifies to (6.5.1).
Conversely suppose that (6.5.1) holds. Then we cannot have (6.5.2) as well. For
if we did, on multiplying (6.5.1) byaand (6.5.2) bybwe would find that

a^2 a 1 +abb 1 = 0 ,−b^2 a 1 +abb 1 = 0 ,

so that(a^2 +b^2 )a 1 =0, and hence as(a,b)=( 0 , 0 ),a 1 =0. Similarly

aba 1 +b^2 b 1 = 0 ,−aba 1 +a^2 b 1 = 0 ,

so thatb 1 =0 as well, giving a contradiction. We now search for a point of intersec-
tion oflandm, and so consider solving for(x,y)the simultaneous equations

ax+by=−c,a 1 x+b 1 y=−c 1.