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Esercizio che chiede di calcolare la distanza e il punto medio tra due punti di coordinate intere assegnate. La distanza \u00e8 sempre un numero naturale.

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Dati i punti $A ( \\var{x_A} ; \\var{y_A} )$ e $B ( \\var{x_B} ; \\var{y_B} )$, determina:

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a) Il segmento $AB$ è l'ipotenusa del triangolo rettangolo di cateti $| x_B - x_A | = | \\simplify[!collectNumbers]{{x_B} - {x_A}} | = \\var{abs(x_B-x_A)}$ e $| y_B - y_A | = |\\simplify[!collectNumbers]{{y_B} - {y_A}}| = \\var{abs(y_B - y_A)}$; usando il teorema di Pitagora otteniamo $A B = \\sqrt{{(x_B - x_A)}^2 + {(y_B - y_A)}^2} = \\sqrt{\\simplify[!collectNumbers]{{({x_B-x_A})}^2 + {({y_B-y_A})}^2}} = \\sqrt{\\var{(x_B-x_A)^2+(y_B-y_A)^2}} = \\var{sqrt((x_B-x_A)^2+(y_B-y_A)^2)}$.

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b) Le coordinate del punto $M$ sono la media tra le coordinate di $A$ e $B$: $x_M = \\frac{x_A + x_B}{2} = \\frac{\\simplify[!collectNumbers]{{x_A}+({x_B})}}{2} = \\var[fractionNumbers]{(x_A+x_B)/2}$, $y_M = \\frac{y_A + y_B}{2} = \\frac{\\simplify[!collectNumbers]{{y_A}+({y_B})}}{2} = \\var[fractionNumbers]{(y_A+y_B)/2}$.

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Ascissa del punto $A$: un numero intero tra $- 6$ e $6$.

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Ordinata del punto $A$: in modo che $A$ e $B$ abbiano una distanza intera (coordinate di $B - A$ in terna pitagorica con la distanza)

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Ascissa del punto $B$: in modo che $A$ e $B$ abbiano una distanza intera (coordinate di $B - A$ in terna pitagorica con la distanza)

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Ordinata del punto $B$: un numero intero tra $- 10$ e $10$.

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Primo generatore di una terna pitagorica

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Secondo generatore di una terna pitagorica

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Costante moltiplicativa della terna pitagorica

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la distanza tra i punti $A$ e $B$:   [[0]]

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le coordinate del punto medio $M$ del segmento $A B$:   $M$([[0]]; [[1]])

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(Per la linea di frazione usa il carattere \"/\".)

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