Question

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

Answer 1

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

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

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

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

Определим площадь по формуле Герона:

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

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

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

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

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

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

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

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

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

Ответ: $S = 8,5$