Дано: А(1;2), B(2;-2), C(6;1)
Найти: α - CD,BM
Пошаговое объяснение:
1. Находим уравнение перпендикуляра CD к стороне АВ.
1) k = ΔY/ΔX = (Аy-Вy)/(Аx-Вx)=(2--2)/(1-2)=-4 - наклон прямой
2) b=Аy-k*Аx=2-(-4)*1=6- сдвиг по оси ОУ
Уравнение Y(АВ) = -4*x+6
Наклон перпендикуляра - k2 = - 1/(-4) = 1/4 = 0.25
Точка C(6,1), наклон - k2 = 0,25
b = Cу - k*Cx = 1 - (0,25)*6 = -0,5
Уравнение высоты - Y(CD) = 0,25*x - 0,5
2. Находим уравнение медианы ВМ.
Находим координаты точки М = (А+С)/2
Мх = (1+6)/2 = 3,5, Му = (2+1)/2 = 1,5.
Находим уравнение медианы ВМ через точку В(2;-2) и М(3,5;1,5)
1) k = ΔY/ΔX = (Мy-Вy)/(Мx-Вx)=(1,5--2)/(3,5-2)=2 1/3 - наклон прямой
2) b=Мy-k*Мx=1,5-(2 1/3 * 3,5= -6 2/3- сдвиг по оси ОУ
Уравнение медианы Y(МВ) = 2 1/3*x - 6 2/3
3. Расчет угла между двумя прямыми. k1 = 2 1/3, k2 = 1/4
tgα = (k2 - k1)/(1 + k1*k2)
k2 - k1 = 2 1/12 = 2,083
1 + k1*k2 = 1 7/12 = 1,583
tg α = 1 6/19 = 1,316
α = 0.921 рад = 52.77° - ОТВЕТ
Пошаговое объяснение:
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