# Points A (1; 2;), B (-3; 0), C (4, -2) are given. Determine the coordinates and modules of the vectors

**Points A (1; 2;), B (-3; 0), C (4, -2) are given. Determine the coordinates and modules of the vectors: AB, AC, BC, AB AC, AB-BC.**

1. The coordinates of the vector with the beginning A (x₁; y₁) and the end B (x₂; y₂) are the numbers a₁ = x₂ – x₁ and a₂ = y₂ – y₁.

AB = (- 3 – 1; 0 – 2) = (- 4; – 2);

AC = (4 – 1; – 2 – 2) = (3; – 4);

BC = (4 – (- 3); – 2 – 0) = (4 + 3; – 2) = (7; – 2).

The sum of vectors a (a₁; a₂) and b (b₁; b₂) is a vector c (c₁; c₂) with coordinates c₁ = a₁ + b₁, c₂ = a₂ + b₂.

AB + AC = (- 4 + 3; – 2 + (- 4)) = (- 1; – 2 – 4) = (- 1; – 6).

The difference of vectors a (a₁; a₂) and b (b₁; b₂) is a vector c (c₁; c₂) with coordinates c₁ = a₁ – b₁, c₂ = a₂ – b₂.

AB – BC = (- 4 – 7; – 2 – (- 2)) = (- 11; 0).

2. The length of the vector a (a₁; a₂) is calculated by the formula:

| a | = √ (a₁² + a₂²).

| AB | = √ ((- 4) ² + (- 2) ²) = √ (16 + 4) = √20 = 2√5;

| AC | = √ (3² + (- 4) ²) = √ (9 + 16) = √25 = 5;

| BC | = √ (7² + (- 2) ²) = √ (49 + 4) = √53;

| AB + AC | = √ ((- 1) ² + (- 6) ²) = √ (1 + 36) = √37;

| AB – BC | = √ ((- 11) ² + 0²) = √ (121 + 0) = √121 = 11.