Plane Geometry - Bruce E. Shapiro

66 SECTION 14. BETWEENNESS(a) To show that f is 1-1, letP = (x 1 , y 1 ), Q = (x 2 , y 2 )and suppose that f(P ) = f(Q) (see definition 6.11). Thenx 1√1 + m2 = f(P ) = f(Q) = x 2√1 + m2Since √ 1 + m 2 ≠ 0 it can be cancelled out, giving x 1 = x 2 . Thusy 1 = mx 1 + b= mx 2 + b= y 2Hence f is 1-1 (P = Q ⇒ f(P ) = f(Q)).(b) To show that f is onto, pick any z ∈ R and definex =z√1 + m2andy = mx + b (14.4)Then P = (x, y) ∈ l because of 14.4 andf(P ) = f(x, y) = x √ 1 + m 2 = zThus (definition 6.12) f is onto R.(c) To verify the distance formula (equation 14.3), let P = (x 1 , y 1 ), Q =(x 2 , y 2 ) ∈ l. Theny 1 = mx 1 + by 2 = mx 2 + b« CC BY-NC-ND 3.0. Revised: 18 Nov 2012