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! :)

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

To show that the points (-3, 1), (3, -1), and (1, 3) are the vertices of an isosceles triangle, we need to prove that at least two sides of the triangle are equal in length.

Let's find the lengths of the sides of the triangle using the distance formula:

The distance between (-3, 1) and (3, -1) is given by:
d1 = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(3 - (-3))^2 + (-1 - 1)^2]
= √[(6)^2 + (-2)^2]
= √[36 + 4]
= √40
= 2√10

The distance between (-3, 1) and (1, 3) is given by:
d2 = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(1 - (-3))^2 + (3 - 1)^2]
= √[(4)^2 + (2)^2]
= √[16 + 4]
= √20
= 2√5

The distance between (3, -1) and (1, 3) is given by:
d3 = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(1 - 3)^2 + (3 - (-1))^2]
= √[(-2)^2 + (4)^2]
= √[4 + 16]
= √20
= 2√5

Now, comparing the distances d1, d2, and d3, we see that d2 = d3 = 2√5, which means that two sides of the triangle are equal in length. Therefore, the given points (-3, 1), (3, -1), and (1, 3) form an isosceles triangle.

To find the area of the triangle, we can use the formula for the area of a triangle given its vertices. Let's call the points A(-3, 1), B(3, -1), and C(1, 3).

The formula for the area of a triangle with vertices A(x1, y1), B(x2, y2), and C(x3, y3) is:
Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Substituting the coordinates of the given points into the formula, we get:
Area = 1/2 * |-3(-1 - 3) + 3(3 - 1) + 1(1 - (-1))|
= 1/2 * |-3(-4) + 3(2) + 1(2)|
= 1/2 * (12 + 6 + 2)
= 1/2 * 20
= 10

Therefore, the area of the isosceles triangle formed by the points (-3, 1), (3, -1), and (1, 3) is 10 square units.