# Quadrilateral ABCD point A (-3; -1) point B (1; 2) point C (5; -1) point D (1; -4) prove that quadrilateral ABCD is a rhombus

For the solution, we use the formula for determining the length of a segment by its coordinates.

d = √ (X2 – X1) ^ 2 + (Y2 – Y1) ^ 2, where d is the length of the segment, X1, X2, Y1, Y2 are the coordinates of the points of the segment.

AB √ (1 – (-3)) ^ 2 + (2 – (-1)) ^ 2 = √16 + 9 = √25 = 5 cm.

AD = √ (1 – (-3)) ^ 2 + (-4 – (-1)) ^ 2 = √16 + 9 = √25 = 5 cm

BC = √ (1 – 5) ^ 2 + (-1 – 2) ^ 2 = √16 + 9 = √25 = 5 cm.

СD = √ (1 – 5) ^ 2 + (-4 – (-1)) ^ 2 = √16 + 9 = √25 = 5 cm.

The lengths of the sides of the quadrangle are equal, it is a square or a rhombus. Let us determine the lengths of the diagonals of the quadrilateral.

AC = √ (5 – (-3)) ^ 2 + (-1 – (-1)) ^ 2 = √64 + 0 = √64 = 8 cm.

ВD = √ (1 – 1) ^ 2 + (-4 – 2) ^ 2 = √0 + 36 = √36 = 6 cm.

Since the diagonals of the quadrilateral are different, ABCD is a rhombus, which is what we had to prove.