Найти S треугольника с координатами A(-2;1) B(1;5) C(-4;4)

Question 1

Question 2

сори, B(4;-4)

Question 3

Если будет ответ с минусом, пишите, мне всё важно.

Question 4

Извините, я решил, но не заметил вашей записи...

Question 5

$A(-2; 1), \ B(1; 5), \ C(-4; 4)$

$AB = \sqrt{(B_{x} - A_{x})^{2} + (B{y} - A_{y})^{2}} = \sqrt{(1 - (-2))^{2} + (5 - 1)^{2}} = \\=\sqrt{3^{2} + 4^{2}} =\sqrt{25} = 5$

$BC = \sqrt{(C_{x} - B_{x})^{2} + (C{y} - B_{y})^{2}} = \sqrt{(-4 - 1)^{2} + (4 - 5)^{2}} = \\=\sqrt{(-5)^{2} + (-1)^{2}} =\sqrt{26}$

$AC = \sqrt{(C_{x} - A_{x})^{2} + (C{y} - A_{y})^{2}} = \sqrt{(-4 - (-2))^{2} + (4 - 1)^{2}} = \\=\sqrt{(-2)^{2} + 3^{2}} =\sqrt{13}$

Определим площадь по формуле Герона:

$p = \dfrac{AB + BC + AC}{2} = \dfrac{5 + \sqrt{26} + \sqrt{13} } {2}$

$S = \sqrt{p(p - AB)(p - BC)(p - AC)} =$

$= \sqrt{\dfrac{5+\sqrt{26} + \sqrt{13}} {2} \bigg(\dfrac{5 + \sqrt{26} +\sqrt{13} }{2} - 5\bigg) \bigg(\dfrac{5+ \sqrt{26} + \sqrt{13}}{2}-\sqrt{26} \bigg) \bigg(\dfrac{5 + \sqrt{26} + \sqrt{13}}{2} - \sqrt{13} \bigg)}$

$= \sqrt{\dfrac{5 +\sqrt{26} + \sqrt{13} } {2} \ \cdotp \dfrac{5 + \sqrt{26} + \sqrt{13}-10} {2} \ \cdotp \dfrac{5 + \sqrt{26}+\sqrt{13}-2\sqrt{26}} {2}\ \cdotp \dfrac{5 + \sqrt{26} +\sqrt{13}-2 \sqrt{13}} {2}} =$

$=\sqrt{\dfrac{5 + \sqrt{26} + \sqrt{13} } {2} \ \cdotp\dfrac{-5 + \sqrt{26} +\sqrt{13}} {2} \ \cdotp \dfrac{5 - \sqrt{26} + \sqrt{13} } {2} \ \cdotp \dfrac{5 + \sqrt{26} - \sqrt{13} } {2} } =$

$= \sqrt{\dfrac{(5 + \sqrt{26} + \sqrt{13})(-5 + \sqrt{26} + \sqrt{13})(5-\sqrt{26} + \sqrt{13})(5 +\sqrt{25}-\sqrt{13})}{16}} =$

$= \sqrt{\dfrac{(5 + \sqrt{26} + \sqrt{13})(13 - (-5 + \sqrt{26})^{2})(5 +\sqrt{25}-\sqrt{13})}{16}} =$

$= \sqrt{\dfrac{((5+\sqrt{26})^{2})(13 - (\sqrt{26} - 5)^{2})}{16}} =\sqrt{\dfrac{(10\sqrt{26} + 38)(-38 + 10 \sqrt{26})}{16}} =$

$= \sqrt{\dfrac{100 \ \cdotp 26 - 1444}{16}} = \sqrt{\dfrac{2600 - 1444}{16}} = \sqrt{ \dfrac{1156}{16}} = \sqrt{ \dfrac{289}{4}} = \dfrac{17}{2} = 8,5$

Ответ: $S = 8,5$