Question

What is the area of a triangle whose vertices are J(-2,1), K(0.3), L(3.-4)?

Answer 1

Check the picture below, so that's the triangle hmmmm a bit non-right-triangle or irregular per se, so let's use Heron's formula on this one, so we'll have to first find the length of each side

`~\hfill \stackrel{\textit{\large distance between 2 points}}{d = √(( x_2- x_1)^2 + ( y_2- y_1)^2)}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ J(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{0}~,~\stackrel{y_2}{3}) ~\hfill JK=√((~~ 0- (-2)~~)^2 + (~~ 3- 1~~)^2) \\\\\\ ~\hfill JK=√(( 2)^2 + ( 2)^2) \implies JK=√( 8 )`

`K(\stackrel{x_1}{0}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{3}~,~\stackrel{y_2}{-4}) ~\hfill KL=√((~~ 3- 0~~)^2 + (~~ -4- 3~~)^2) \\\\\\ ~\hfill KL=√(( 3)^2 + ( -7)^2) \implies KL=√( 58 ) \\\\\\ L(\stackrel{x_1}{3}~,~\stackrel{y_1}{-4})\qquad J(\stackrel{x_2}{-2}~,~\stackrel{y_2}{1}) ~\hfill LJ=√((~~ -2- 3~~)^2 + (~~ 1- (-4)~~)^2) \\\\\\ ~\hfill LJ=√(( -5)^2 + (5)^2) \implies LJ=√( 50 )`

now let's use those three lengths for Heron's

`\qquad \textit{Heron's area formula} \\\\ A=√(s(s-a)(s-b)(s-c))\qquad \begin{cases} s=(a+b+c)/(2)\\[-0.5em] \hrulefill\\ a=√(8)\\ b=√(58)\\ c=√(50)\\ s=(√(8)+√(58)+√(50))/(2)\\\\ \qquad (√(58)+7√(2))/(2) \end{cases}`

`A=\sqrt{(√(58)+7√(2))/(2)\left((√(58)+7√(2))/(2)-√(8) \right)\left((√(58)+7√(2))/(2)-√(58) \right)\left((√(58)+7√(2))/(2)-√(50) \right)} \\\\\\ ~\hfill {\Large \begin{array}{llll} A=10 \end{array}}~\hfill`