#separator:tab #html:false #notetype:Basic #deck:Math::Distance formula #tags:math precalculus distance-formula #columns:Front Back Tags #tags column:3 State the distance formula. d(P1, P2) = the square root of (x2 - x1)^2 + (y2 - y1)^2. formula The distance formula in words? Subtract the x's and square, subtract the y's and square, add, take the square root. formula Which theorem is the distance formula? The Pythagorean theorem, with the legs read off the coordinates. why Distance formula: where is the right angle? At P3 = (x2, y1), where a horizontal line through P1 meets a vertical line through P2. why Distance formula: what are the two legs? The horizontal leg |x2 - x1| and the vertical leg |y2 - y1|. why Why does the order of the two points not matter? Swapping them flips the sign of each difference, and squaring erases the sign. why Why does the distance formula need no absolute values? The legs are |x2 - x1| and |y2 - y1|, but |t|^2 = t^2, so the squares do the job. why What is the square root of t^2? |t|, not t. The square root is never negative. why Horizontal points (x1, y) and (x2, y): what does the formula give? |x2 - x1|, the distance on a number line. why How do you compare two distances without taking square roots? Compare their squares: for non-negative numbers, squaring keeps the order. why When is d(P1, P2) = 0? Only when P1 = P2: a sum of two squares is 0 only when both are 0. why Distance from (0, 0) to (a, b)? The square root of a^2 + b^2. formula Would the distance formula be right on axes that are not perpendicular? No. It needs the right angle between horizontal and vertical. why Distance from (1, 3) to (5, 6)? 5. Legs 4 and 3, corner at (5, 3). problems Distance from (-4, 5) to (3, 2)? The square root of 58, about 7.62: 49 + 9. problems Distance from (0, 0) to (1, 1)? The square root of 2, about 1.414. problems Distance from (-2, 4) to (5, 4)? 7. Same y, so |5 - (-2)|. problems Simplify the square root of 20. 2 times the square root of 5, since 20 = 4 times 5. problems A = (-2, 1), B = (2, 3), C = (3, 1): right triangle? Yes. AB^2 = 20, BC^2 = 5, AC^2 = 25 = 20 + 5. The right angle is at B. problems Closer to the origin: (5, -1) or (-3, 4)? (-3, 4): 25 against 26, comparing squares. problems Points on the x-axis at distance 5 from (1, 3)? (5, 0) and (-3, 0): (x - 1)^2 + 9 = 25. problems How can distances show three points lie on one line? The two short distances add up exactly to the long one. why What is the triangle inequality in coordinates? d(P, R) is at most d(P, Q) + d(Q, R), equal only when Q is on the segment PR. why Trap: the square root of a^2 + b^2 equals a + b? No. The square root of 9 + 16 is 5, not 7. trap Trap: (x2 - x1)^2 + (y2 - y1)^2 equals (x2 - x1 + y2 - y1)^2? No. From (0, 0) to (3, 4): 9 + 16 = 25, but 7^2 = 49. trap Trap: from (3, -1) to (3, -7) the distance is -6? No. A distance is never negative: 6. trap Trap: (3 - -4)^2 = (3 - 4)^2 = 1? No. 3 - (-4) = 7, so the square is 49. Put negative coordinates in brackets. trap Trap: the subscript in x2 means x squared? No. x2 is the x-coordinate of the second point; a square is written x^2. trap Trap: -3^2 = 9? No. -3^2 = -9; the square applies before the minus. (-3)^2 = 9. trap Trap: to compare two distances you must compute both square roots? No. Compare the squares: 26 against 25 already decides it. trap