Стороны треугольника лежат на прямых x+5у–7=0, 3x–2y–4=0, 7x+y+19=0. Вычислить его площадь S.
Находим координаты вершин треугольника как точки пересечения заданных прямых.
3x–2y–4=0, 3x–2y–4=0,
7x+y+19=0 |x2 = 14x+2y+38=0
17x + 34 = 0, x = -34/17 = -2.
y = (3/2)*x - (4/2) = y = (3/2)*(-2) - (4/2) = -3 - 2 = -5.
Точка А(-2; -5).
x+5у–7=0, |x-7 = -7x-35y+49=0
7x+y+19=0, 7x+y+19=0
-34y+68 = 0, y = -68/-34 = 2.
x = 7 - 5y = 7 - 5*2 = -3.
Точка В(-3; 2).
x+5у–7=0, |x(-3) = -3x-15y+21 = 0
3x–2y–4=0 3x–2y–4 = 0
-17y+17 = 0, y = -17/-17 = 1.
x = 7 - 5y = 7 - 5*1 = 2.
Точка С(2; 1).
Найдем вектора по координатам точек:
AB = {Bx - Ax; By - Ay; Bz - Az} = {-3 - (-2); 2 - (-5); 0 - 0} = {-1; 7; 0}
AC = {Cx - Ax; Cy - Ay; Cz - Az} = {2 - (-2); 1 - (-5); 0 - 0} = {4; 6; 0}
S = (1/2) |AB × AC|
Найдем векторное произведение векторов:
c = AB × AC
AB × AC =
i j k
ABx ABy ABz
ACx ACy ACz
=
i j k
-1 7 0
4 6 0
= i (7·0 - 0·6) - j ((-1)·0 - 0·4) + k ((-1)·6 - 7·4) =
= i (0 - 0) - j (0 - 0) + k (-6 - 28) = {0; 0; -34}
Найдем модуль вектора:
|c| = √(cx^2 + cy^2 + cz^2) = √(0^2 + 0^2 + (-34)^2) = √(0 + 0 + 1156) = √1156 = 34
Найдем площадь треугольника:
S = (1/2)* 34 = 17 .
Пошаговое объяснение:
1.
1) 11/15 : 3/8 = 11/15 * 8/3 =
88/45 = 1. 43/45
2) 6/35 : 18/25 = 6/35 * 25/18 = 1/7 * 5/3 = 5/21
3) 12/55 : 48/77 = 12/55 * 77/48 = 1/5 * 7/4 = 7/20
4) 21/40 : 3/4 = 21/40 * 4/3 = 7/10 * 1/1 = 7/10
5) 27/50 : 9/25 = 27/50 * 25/9 = 3/2 * 1/1 = 3/2 = 1 1/2
6) 63/64 : 45/56 = 63/64 * 56/45 = 7/8 * 7/5 = 49/40 = 1 9/40
7) 5/8 : 5/32 = 5/8 * 32/5 = 32/8 = 4
8) 14/55 : 1/5 = 14/55 * 5/1 = 14/11 * 1/1 = 14/11 = 1 3/11
2.
1) 6 : 7/9 = 6 * 9/7 = 54/7 = 7 5/7
2) 16 : 4/11 = 16 * 11/4 = 176/4 = 44
3) 13 : 26/29 = 13 * 29/26 = 377/26 = 14 13/26 = 14 1/2
4) 7/9 : 5 = 7/9 * 1/5 = 7/45
5) 9/16 : 6 = 9/16 * 1/6 = 9/96 = 3/32
6) 1 5/9 : 1 8/27 = 14/9 * 27/35 = 2/1 * 3/5 = 6/5 =
1 1/5
7) 2 10/13 : 3 3/26 = 36/13 * 26/81 = 4/1 * 2/9 =
8/9
8) 2 4/7 : 1 1/35 = 18/7 * 35/36 = 1/1 * 5/2 = 5/2 =
2 1/2
