Prove that the quadrilateral ABCD is a rectangle, find its area if A (14; 1), B (23; 4), C (21; 10) and D (12; 7).

Given: quadrilateral ABCD.

It is known. A (14; 1), B (23; 4), C (21; 10) and D (12; 7).

To find. Square.

Check. Is it a rectangle.

The length of the segment specified by the coordinates of its ends is determined by the formula:

a = √ ((x2 – x1) ^ 2 + (y2 – y1) ^ 2).

Find the lengths of the sides of the quadrangle:
AB = √ ((23 – 14) ^ 2 + (4 – 1) ^ 2) = √90.
AD = √ ((12 – 14) ^ 2 + (7 – 1) ^ 2) = √40.
BC = √ ((23 – 21) ^ 2 + (4 – 10)) ^ 2 = √40.
CD = √ ((12 – 21) ^ 2 + (7 – 10) ^ 2) = √90.
The sides are pairwise equal.
Let us find the coordinates of the vectors AB {9; 3}, BC {2; -6}, CD {-9; -3}, AD {-2; 6}.
Let’s find the scalar product of vectors, if it is equal to zero, then they are perpendicular.
AB * BC = 18 – 18 = 0. -> Angle between sides AB and BC = 90 °.

In general, if the sides are equal in pairs and one of the corners in the quadrilateral is 90 °, then this is a rectangle.

Area S = | AB | * | BC | = √90 * √40 = √3600 = 60.

Answer. Rectangle, area – 60 sq.