A(-3;2),B(2;3),C(1;-2),D(-4;-3) show that ABCD is a rhombus and AC & BD bisect each other

To show that ABCD is a rhombus, we need to prove that all four sides are equal in length.

Let's calculate the distance between each pair of consecutive vertices:

1. The distance between A(-3, 2) and B(2, 3) can be found using the distance formula:
d_AB = sqrt((x2 - x1)^2 + (y2 - y1)^2)
d_AB = sqrt((2 - (-3))^2 + (3 - 2)^2)
= sqrt(5^2 + 1^2)
= sqrt(25 + 1)
= sqrt(26)

2. The distance between B(2, 3) and C(1, -2) can be found using the distance formula:
d_BC = sqrt((x2 - x1)^2 + (y2 - y1)^2)
d_BC = sqrt((1 - 2)^2 + (-2 - 3)^2)
= sqrt((-1)^2 + (-5)^2)
= sqrt(1 + 25)
= sqrt(26)

3. The distance between C(1, -2) and D(-4, -3) can be found using the distance formula:
d_CD = sqrt((x2 - x1)^2 + (y2 - y1)^2)
d_CD = sqrt((-4 - 1)^2 + (-3 - (-2))^2)
= sqrt((-5)^2 + (-1)^2)
= sqrt(25 + 1)
= sqrt(26)

4. The distance between D(-4, -3) and A(-3, 2) can be found using the distance formula:
d_DA = sqrt((x2 - x1)^2 + (y2 - y1)^2)
d_DA = sqrt((-3 - (-4))^2 + (2 - (-3))^2)
= sqrt(1^2 + 5^2)
= sqrt(1 + 25)
= sqrt(26)

We have calculated the lengths of all four sides and found that d_AB = d_BC = d_CD = d_DA = sqrt(26). Therefore, ABCD is a rhombus.

Next, let's prove that the diagonals AC and BD bisect each other.

The midpoint of a line segment with coordinates (x1, y1) and (x2, y2) can be found using the midpoint formula:

Midpoint = ((x1 + x2) / 2, (y1 + y2) / 2)

1. The midpoint of AC can be found as:
Midpoint_AC = ((-3 + 1) / 2, (2 + (-2)) / 2)
= (-1 / 2, 0)
= (-0.5, 0)

2. The midpoint of BD can be found as:
Midpoint_BD = ((2 + (-4)) / 2, (3 + (-3)) / 2)
= (-2 / 2, 0)
= (-1, 0)

We can see that both midpoints have the same x-coordinate of -1 and y-coordinate of 0. Therefore, AC and BD bisect each other.

Hence, we have shown that ABCD is a rhombus and its diagonals AC and BD bisect each other.