Предмет: Геометрия, автор: NeymarJunior1925

Напиши уравнение прямой ax+by+c=0, все точки которой находятся на равных расстояниях от точек A(3;4) и B(8;9) .

Ответы

Автор ответа: a87086828
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Ответ:

Объяснение:

Прямая которая находится на равном расстоянии от двух точек А и В является серединным перпендикуляром к отрезку АВ.

Координаты середины отрезка АВ (М) находятся по формулам:

M_x=\frac{A_x+B_x}{2}=\frac{3+8}{2}=5.5\\ M_y=\frac{A_y+B_y}{2}=\frac{4+9}{2}=6.5

M(5.5;6.5)

Теперь найдем угловой коэффициент прямой АВ(k):

k=\frac{B_y-A_y}{B_x-A_x}=\frac{9-4}{8-3}=1

Угловой коэффициент перпендикулярный прямой равен обратному отрицательному значению углового коэффициента исходной прямой, то есть k'=-1.

Теперь мы можем записать уравнение прямой в отрезке, проходящем через точку М и перпендикулярным АВ:

y-M_y=k'(x-M_x)

Подставим значения:

y-6.5=-1*(x-5.5) или раскрыв скобки и приведя подобные слагаемые:

x+y-12=0

Таким образом, уравнение прямой, все точки которой находятся на равных расстояниях от точек A(3;4) и B(8;9), является x + y - 12 = 0.

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