# Find the area of a parallelogram whose vertices have coordinates (-2; 1.5) (3; 2.5) (4; -1) (-1; -2)

On the Cartesian coordinate plane Oxy, mark the points A (-2; 1.5), B (3; 2.5), C (4; -1) and D (-1; -2). The task requires to find the area of ​​the quadrilateral ABCD. We intentionally replaced the word “parallelogram” with the word “quadrangle”, because first we have to prove that the figure (quadrilateral) ABCD is a parallelogram. For this purpose, we will compose the equations of all sides of the quadrilateral ABCD.
For side AB we have: (x – (-2)) / (3 – (-2)) = (y – 1.5) / (2.5 – 1.5), whence y = 0.2 * x + 1.9. The next side of the aircraft. Similarly, (x – 3) / (4 – 3) = (y – 2.5) / (-1 – 2.5) or y = -3.5 * x + 13. Next, the CD side. We calculate: (x – 4) / (-1 – 4) = (y – (-1)) / (-2 – (-1)), whence y = 0.2 * x – 1.8. Finally, the equation for the side DA is: (x – (-1)) / (-2 – (-1)) = (y – (-2)) / (1.5 – (-2)) or y = – 3.5 * x – 5.5. The obtained equations of straight lines with a slope confirm that: AB || СD and ВС || DA, since parallel straight lines have equations with equal slopes. This indicates that the quadrilateral is indeed a parallelogram.
Now let’s start calculating the area S of the parallelogram ABCD. There are a number of ways to calculate the area of ​​a parallelogram. We will calculate the area of ​​this parallelogram ABCD by the base CD and the height (which we denote by BE) dropped to the side of CD from the top B (3; 2.5). Let’s calculate the length of the side СD according to the formula of the distance between two points: СD = √ (-1 – 4) ² + (-2 – (-1) ²)) = √ ((- 5) ² + (-1) ²) = √ (26). Since the equation of the side СD is known, the length of the height BE will be found by the formula for the distance from a point to a straight line. First, we rewrite the equation of the straight line CD in the form x – 5 * y – 9 = 0. We have: BE = | 3 – 5 * 2.5 – 9 | / √ (1² + (-5) ²) = 18.5 / √ (26). Then S = СD * BE = √ (26) * (18.5 / √ (26)) = 18.5. 