can anyone help

A triangle has coordinates A(1,1),B(4,2),C(3,5).  What is the area of triangle ABC?

AB = BC = sqrt( 10)

Angle ABC = 90 degrees

area = (AB * BC) / 2 = 5 

A triangle has coordinates A(1,1),B(4,2),C(3,5). What is the area of triangle ABC?

Hello Guest!

$y=m(x-x_1)+y_1$

$m_a=\frac{5-2}{3-4}=-3\\ m_b=\frac{5-1}{3-1}=2\\ m_c=\frac{2-1}{4-1}=\frac{1}{3}$

$y_a=-3(x-4)+2=-3x+14\\ y_b=2(x-1)+1=2x-1\\ y_c=\frac{1}{3}(x-1)+1=\frac{1}{3}x+\frac{2}{3}$

$A_{ABC}=\int _3^4(-3x+14)+\int_1^3(2x-1)-\int_1^4(\frac{1}{3}x+\frac{2}{3})$

$A_{ABC}=|-\frac{3}{2}x^2+14x|_3^4+|x^2-x|_1^3-|\frac{1}{6}x^2+\frac{2}{3}x|_1^4$

$A_{ABC}=(-24+56+13.5-42)+(9-3-1+1)$

                                                        $-(\frac{16}{6}+\frac{8}{3}-\frac{1}{6}-\frac{2}{3})=5$

$\large A_{ABC}=5$
  !

A triangle has coordinates A(1,1),B(4,2),C(3,5).  What is the area of triangle ABC?

Area = {sqrt[10 - (√20/2)² ]} * √20 / 2 = 5 u²