Question

find the area of the triangle whose vertices (-5,-1) (3,-5) and (5,2)

Answer 1

let,
ΔABC a tringale which vartex point A(-5,-1), B(3,-5) and C(5,2) Again Let AB=a,BC=b and AC=c
Distance of AB=a=sqrt[(-5-3)²+({-1-(-5)}²]
a=sqrt{(-8)²+4²}=sqrt( 64+16)= √80=4√5
Distance of BC=b=sqrt{(3-5)²+(-5-2)²}
b=sqrt{ (-2)²+(-7)²}=sqrt( 4+49)=√53
Distance of AC=c=sqrt{(-5-5)²+(-1-2)²}
c=sqrt {(-10)²+(-3)²}=sqrt(100+9)=√109

Half-parameter (s) =(a+b+c)/2=(4√5+√53+√109)/2=(8.9+7.3+10.4)2
s=26.6/2=13.3
We got
a=4√5=8.9 ,b=√53=7.3 and c=√109=10.4
we know that hiron Formula
areaΔ=sqrt {s(s-a)(s-b)(s-c)}
therefore
areaΔABC =sqrt{13.3(13.3-8.9)(13.3-7.3)(13.3-10.4)}
=sqrt {13.3(4.4)(6)(2.9)}
area ΔABC=sqrt(1018.25)=31.9 unit
:. area of ΔABC=31.9 unit
I hope that it's your answer..