in the xy coordinate plane point b (-3,2) and point d (3,-4)are opposite vertices of square abcd if point h is the midpoint of line segment bc what is the area of triangle ahd ?

If you plot the points, you can see that a = (-3,-4) and h = (0,2).

So, the base of ahd is the line ad with length 6. The height of the triangle is also 6.

So, A(ahd) = 1/2 * 6 * 6 = 18

To find the area of triangle AHD, we first need to determine the coordinates of point H. We can do this by finding the midpoint of line segment BC.

The coordinates of point B are (-3, 2) and the coordinates of point C are unknown. However, since B and C are opposite vertices of a square, the x-coordinate of C will be the opposite of the y-coordinate of B, and the y-coordinate of C will be the opposite of the x-coordinate of B.

Therefore, the coordinates of point C will be (2, -3).

To find the midpoint of BC, we average the x-coordinates and the y-coordinates of B and C:

Midpoint x-coordinate = (x-coordinate of B + x-coordinate of C) / 2
= (-3 + 2) / 2
= -1/2

Midpoint y-coordinate = (y-coordinate of B + y-coordinate of C) / 2
= (2 + (-3)) / 2
= -1/2

Thus, the coordinates of point H are (-1/2, -1/2).

Now, we have the coordinates of points A, H, and D. The coordinates of point A are (0, 0) and the coordinates of point D are (3, -4).

Using the formula for the area of a triangle, which is 0.5 times the base times the height, we can calculate the area of triangle AHD.

Base = distance between points D and A = √((x2 - x1)^2 + (y2 - y1)^2)
= √((3 - 0)^2 + (-4 - 0)^2)
= √(9 + 16)
= √25
= 5

Height = y-coordinate of point H = -1/2

Area of triangle AHD = 0.5 * base * height
= 0.5 * 5 * (-1/2)
= 0.5 * (-5/2)
= -2.5

Therefore, the area of triangle AHD is -2.5 square units.