# 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.

Answer: 18.5.