The vertices of a quadrilateral are A(-2,3), B(5,4), C(4, -3) and D(-3, -4).

Prove that ABCD is a rhombus. Find the area of ABCD.

Formula to find length:

d = √(x2 - x1)^2 + (y2-y1)^2

Please show working out, thanks.

To prove that ABCD is a rhombus, we need to show that all four sides of the quadrilateral are equal in length. Here's how you can find the length of each side:

1. Side AB:
Using the distance formula, d = √((x2 - x1)^2 + (y2 - y1)^2), we can find the length of AB.
dAB = √((5 - (-2))^2 + (4 - 3)^2)
= √(7^2 + 1^2)
= √(49 + 1)
= √50
= 5√2

2. Side BC:
Using the distance formula, d = √((x2 - x1)^2 + (y2 - y1)^2), we can find the length of BC.
dBC = √((4 - 5)^2 + (-3 - 4)^2)
= √((-1)^2 + (-7)^2)
= √(1 + 49)
= √50
= 5√2

3. Side CD:
Using the distance formula, d = √((x2 - x1)^2 + (y2 - y1)^2), we can find the length of CD.
dCD = √((-3 - 4)^2 + (-4 - (-3))^2)
= √((-7)^2 + (-1)^2)
= √(49 + 1)
= √50
= 5√2

4. Side AD:
Using the distance formula, d = √((x2 - x1)^2 + (y2 - y1)^2), we can find the length of AD.
dAD = √((-3 - (-2))^2 + (-4 - 3)^2)
= √((-1)^2 + (-7)^2)
= √(1 + 49)
= √50
= 5√2

Since all four sides AB, BC, CD, and AD have the same length of 5√2, we can conclude that ABCD is a rhombus.

To find the area of ABCD, we can use the formula for the area of a rhombus:

Area = (d1 * d2) / 2

where d1 and d2 are the lengths of the diagonals.

1. Diagonal AC:
Using the distance formula, d = √((x2 - x1)^2 + (y2 - y1)^2), we can find the length of AC.
dAC = √((4 - (-2))^2 + (-3 - 3)^2)
= √((6)^2 + (-6)^2)
= √(36 + 36)
= √72
= 6√2

2. Diagonal BD:
Using the distance formula, d = √((x2 - x1)^2 + (y2 - y1)^2), we can find the length of BD.
dBD = √((5 - (-3))^2 + (4 - (-4))^2)
= √((8)^2 + (8)^2)
= √(64 + 64)
= √128
= 8√2

Now, we can calculate the area using the formula:

Area = (d1 * d2) / 2
= (6√2 * 8√2) / 2
= (6 * 8 * √2 * √2) / 2
= (48 * 2) / 2
= 48 square units

Therefore, the area of ABCD is 48 square units.

huh? You can't apply the distance formula?

AB = √((5+2)^2+(4-3)^2) = √(49+1) = √50
CD = √((-3-4)^2+(-4+3)^2) = √(49+1) = √50
So, AB=CD

Now do the same steps for AD and BC.