Предмет: Математика, автор: rafikchanel

Даны вершины треугольника АВС A(6,0) B(30,-7) C(12,17). Найти уравнения высоты АД и медианы АМ.

Ответы

Автор ответа: dnepr1
0
Даны вершины треугольника АВС: A(6; 0), B(30; -7), C(12; 17).
Найти уравнения: 1) высоты АД и 2) медианы АМ.

2) Находим координаты точки М - середины стороны ВС.
М((30+12)/2=21; (-7+17)/2=5) = (21; 5).
Теперь по координатам двух точек 
A(6; 0) и М(21; 5) определяем уравнение прямой (медианы АМ), проходящей через эти точки.
AM: (x-6)/(21-6) = (y-0)/(5-0).
AM: (x-6)/15 = y/5   это каноническое уравнение.
Если сократить на 3 и привести подобные, то получим уравнение общего вида:
х - 6 = 3у   или х - 3у - 6 = 0.
Если выразить относительно у, то получим уравнение с угловым коэффициентом:
у =(1/3)х - 2.

1) Определяем уравнение стороны ВС.
B(30; -7), C(12; 17).
(х-30)/(12-30) = (у+7)/)17+7),
(х-30)/(-18) = (у+7)/24.
Или в общем виде: 4х + 3у - 99 = 0.
Или с коэффициентом у = (-4/3)х + 33.
Уравнение перпендикулярной прямой АД имеет угловой коэффициент
к(АД) = -1/(к(ВС)) = -1/(-4/3) = 3/4.
Тогда уравнение АД: у = (3/4)х + в.
Для определения параметра в подставим в уравнение координаты точки А: 0 = (3/4)*6 + в. Отсюда в = -18/4 = -9/2.
Получаем уравнение:
АД: у = (3/4)х - (9/2),
или в общем виде 3х - 4у - 18 = 0.
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