Complex numbers and polynomials - University College Cork

3.2 Distance formulaVarious candidates present themselves that qualify as a ‘distance’ between a pair ofpoints P = (x 1 , y 1 ), Q = (x 2 , y 2 ). The usual one—which you’ll recognise—is givenby|P Q| ≡ d 2 (P, Q) = √ (x 2 − x 1 ) 2 + (y 2 − y 1 ) 2 .Another one—called the taxi-cab metric—is given byAn even simpler one is given byd 1 (P, Q) = |x 2 − x 1 | + |y 2 − y 1 |.d 0 (P, Q) ={ 1, if P ≠ Q,0, if P = Q.If, for the moment, d is any one of these, L is a straight line, and P 0 = (x 0 , y 0 ) isany point, is there a point R in L which is ‘nearest’ to P 0 ? If so, is it unique? Inother words, does there exist a point R ∈ L such thatd(P 0 , R) ≤ d(P 0 , X), ∀X ∈ L,and, if so, is it unique?This is an example of what’s called an Existence and Uniqueness Problem, andpossibly the first one of its kind that is encountered in second level mathematics. Ifd = d 2 , the answer to both questions is in the affirmative, as you know: the foot ofthe perpendicular from P 0 onto L is the point that is nearest to P 0 in this metric.If L = {(x, y) : ax + by + c = 0, a 2 + b 2 > 0}, thenR = ( ac − aby 0 + b 2 x 0a 2 + b 2 , bc + a2 y 0 − abx 0a 2 + b 2 ),and|P 0 R| = |ax 0 + by 0 − c|√a2 + b 2 .However, by contrast, while the answer to the first question is still in the affirmativeif we measure distance using either d 0 or d 1 , we lose uniqueness. For instance, if weuse d 0 , then the distance between every point in L and P 0 is 1, unless P 0 ∈ L, inwhich case the distance between them is 0, and R = P 0 .Exercise 3. Work out a solution to the problem when d 1 is used.Exercise 4. Show that if P, Q, R are three points in the plane, thend 1 (P, Q) ≤ d 1 (P, R) + d 1 (R, Q).6