Find the area of the triangle whose vertices are (2,3),(3,4),(-2,-4).

easiest way:

list the points in a column, repeat the one you started with
2 3
-2 -4
3 4
2 3

area = (1/2)[ sum of downproducts - sum of up-products]
= (1/2)[-8 -8 + 9 - (-6 -12 +8) ]
= (1/2)( -7 + 10]
= 3/2

or

label the points A(2,3) , B(3,4) and C(-2,-4)
BC = √(25 + 64) = √89

equation of BC:
slope BC = -8/-5 = 8/5
using the point (3,4)
y-4 = (8/5)(x-3)
5y - 20 = 8x - 24
8x - 5y - 4 = 0

consider BC as the base and we need the height from A to BC
distance = |8(2) -5(3)-4\/√(64+25) = 3/√89
area = (1/2)(√89)(3/√89)
= 3/2

To find the area of a triangle with given vertices, you can use the shoelace formula.

1. First, write down the given vertices:
A(2, 3)
B(3, 4)
C(-2, -4)

2. Arrange the vertices in a counterclockwise order. Start by writing the coordinates of point A, then point B, and finally point C:
A(2, 3)
B(3, 4)
C(-2, -4)

3. Now, write down the coordinates of point A again as the last point in the list:
A(2, 3)
B(3, 4)
C(-2, -4)
A(2, 3)

4. Multiply the corresponding coordinates of each pair of adjacent points (x1y2, x2y1), and then subtract the sum of these products for the points in the second column from the sum of the products for the points in the first column.

(2*4 + 3*(-4) + (-2)*3) - (3*3 + (-2)*(-4) + 2*4)
(8 - 12 - 6) - (9 + 8 + 8)
(-10) - (25)
-10 - 25
-35

5. Take the absolute value of the result obtained in step 4. This will give you the area of the triangle.
| -35 |
35

Therefore, the area of the triangle with vertices (2, 3), (3, 4), and (-2, -4) is 35 square units.