Plane Geometry - Bruce E. Shapiro

SECTION 14. BETWEENNESS 67henceP Q = d(P, Q)»= (x 2 − x 1 ) 2 + (y 2 − y 1 ) 2»= (x 2 − x 1 ) 2 + (mx 2 + b − mx 1 − b) 2»= (x 2 − x 1 ) 2 + (m(x 2 − x 1 )) 2»= (1 + m 2 )(x 2 − x 1 ) 2= √ 1 + m 2 |x 2 − x 1 |∣√ √ = ∣x 2 1 + m2 − x 1 1 + m2∣= |y 2 − y 1 |= |f(P ) − f(Q)|Thus if l is not a vertical line, f is a coordinate function.Now suppose that l is a vertical line with equation x = a and define f :l ↦→ R by f(a, y) = y.(a) Let P = (a, y 1 ), Q = (a, y 2 ) ∈ l where P ≠ Q, hence y 1 ≠ y 2 andf(P ) = y 1 ≠ y 2 = f(Q)which shows that f is 1-1 (P ≠ Q ⇒ f(P ) ≠ f(Q)).(b) Let y ∈ R be any number. Then P = (a, y) ∈ l and f(P ) = y. Hence lis onto.(c) Let P = (a, y 1 ) and Q = (a, y 2 ). ThenP Q =»(a − a) 2 + (y 2 − y 1 ) 2= |y 2 − y 1 |= |f(Q) − f(P )|Example 14.3 (The “Taxicab” Metric) Let P = (x 1 , y 1 ), Q = (x 2 , y 2 ).Thend(P, Q) = |x 1 − x 2 | + |y 1 − y 2 |A coordinate function for the taxicab metric is given byf(x, y) =®x(1 + |m|), if l is not vertical;y, if l is verticalRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.