To prove that ABCD is a parallelogram, we need to show that both pairs of opposite sides are parallel.
First, let's find the slopes of the line segments AB, BC, CD, and DA.
The slope of AB = (y2 - y1) / (x2 - x1) = (3 - 0) / (4 - 0) = 3/4.
The slope of BC = (y2 - y1) / (x2 - x1) = (7 - 3) / (1 - 4) = 4/(-3) = -4/3.
The slope of CD = (y2 - y1) / (x2 - x1) = (4 - 7) / (-3 - 1) = -3/(-4) = 3/4.
The slope of DA = (y2 - y1) / (x2 - x1) = (0 - 4) / (0 - (-3)) = -4/3.
We can see that the slopes of AB and CD are equal (3/4), and the slopes of BC and DA are equal (-4/3). Therefore, both pairs of opposite sides are parallel, and we have proven that ABCD is a parallelogram.
To find the area of the quadrilateral, we can divide it into two triangles: ABC and CDA. The area of a triangle can be calculated using the Shoelace Formula or by applying the formula: Area = (1/2) * base * height.
For triangle ABC:
Base = distance between A and B = √[(4 - 0)^2 + (3 - 0)^2] = √(16 + 9) = √25 = 5
Height = distance between A and C = √[(1 - 0)^2 + (7 - 0)^2] = √(1 + 49) = √50 = 5√2
Area(ABC) = (1/2) * 5 * 5√2 = 25√2
For triangle CDA:
Base = distance between C and D = √[(1 - (-3))^2 + (7 - 4)^2] = √(16 + 9) = √25 = 5
Height = distance between C and A = √[(0 - 1)^2 + (0 - 7)^2] = √(1 + 49) = √50 = 5√2
Area(CDA) = (1/2) * 5 * 5√2 = 25√2
The total area of the quadrilateral ABCD = Area(ABC) + Area(CDA) = 25√2 + 25√2 = 50√2
To find the perimeter of the quadrilateral, we need to calculate the sum of the lengths of all four sides.
AB = √[(4 - 0)^2 + (3 - 0)^2] = √(16 + 9) = √25 = 5
BC = √[(1 - 4)^2 + (7 - 3)^2] = √(9 + 16) = √25 = 5
CD = √[(-3 - 1)^2 + (4 - 7)^2] = √(16 + 9) = √25 = 5
DA = √[(0 - (-3))^2 + (0 - 4)^2] = √(9 + 16)