show that (-3, 1) and (3, -1) and (1, 3) are the vertices of an isosceles triangle. find its area. please include explanation. this is really frustrating please help. thanks alot! :)

Answer 1

label the points
A(-3,1) , B(3,-1) and C(1,3)

AB = √(6^2 + (-2)^2) = √40
BC= √(2^2 + 4^2) = √20
AC = √(4^2 + 2^2) = √20

Two of the sides are equal so the triangle is isosceles.

here is a quick way to find the area if you are given the 3 points.
list the coordinates in a column, repeating the first point

-3 1
3 -1
1 3
-3 1

area = (1/2) | downproducts - upproducts|
= (1/2) |3+9+1 - (3-1-9)|
= (1/2)|13 +7|
= 10

or use Heron's Formula

Area = √(s(s-a)(s-b)(s-c)) , where a, b, and c are the sides and s = (1/2) the perimeter

s = (√40 + 2√20)/2 = (2√10 + 4√5)/2 = √10 + 2√5
s-a = √10 + 2√5 - 2√5 = √10
s-b = √10 + 2√5 - 2√5 = √10
s-c = √10 + 2√5 - 2√10 = 2√5 - √10
area = √(√10+2√5)(√10)(√10)(2√5-√10)
= √10(20-10)
= 10