find exact area of the triangle ABC?Points: here's how: -...

src="/jax/includes/tinymce/jscripts/tiny_mce/plugins/asciisvg/js/d.svg"
sscr="-7.5,7.5,-2,8,1,1,1,1,1,300,200,param,5t-3,3t+4,0,0,0,1,black,1,none,param,3t+2,-6t+7,0,0,0,1,black,1,none,param,8t-3,-3t+4,0,0,0,1,black,1,none"/>

class="AM">`c=|AB| = sqrt((-3 - 2)^2+(4 - 7)^2) = sqrt(25+9) = sqrt(34)`

`a=|BC| = sqrt((2 - 5)^2+(7-1)^2) = sqrt(9+36) =
sqrt(45)=3sqrt(5)`
`b=|AC| = sqrt((-3 -
5)^2+(4-1)^2)) = sqrt(64+9) = sqrt(73)`

We can
use Heron's formula for area

`A =
sqrt(s(s-a)(s-b)(s-c))` where `s =
(a+b+c)/2`

`s =
(sqrt(34)+3sqrt(5)+sqrt(73))/2`
class="AM">`s-a=(sqrt(34)+3sqrt(5)+sqrt(73))/2 -
3sqrt(5)=(sqrt(34)+sqrt(73)-3sqrt(5))/2`

class="AM">`s-b=(sqrt(34)+3sqrt(5)-sqrt(73))/2`

class="AM">`s-c=(3sqrt(5)+sqrt(73)-sqrt(34))/2`

so

class="AM">`A=sqrt(((sqrt(34)+3sqrt(5)+sqrt(73))/2)((sqrt(34)+sqrt(73)-3sqrt(5))/2)((sqrt(34)+3sqrt(5)-sqrt(73))/2)((3sqrt(5)+sqrt(73)-sqrt(34))/2))` class="AM">`or`

class="AM">`A=1/4sqrt((sqrt(34)+3sqrt(5)+sqrt(73))(sqrt(34)+sqrt(73)-3sqrt(5))(sqrt(34)+3sqrt(5)-sqrt(73))(3sqrt(5)+sqrt(73)-sqrt(34))`