A triangle with vertices at points M (2; 3), N (-4; 6) and K (5; -1), define the cosine of the angle M.

Let’s use the formula for the distance between two points A and B on the coordinate plane with coordinates A (x1; y1) and B (x2; y2):

| AB | = √ ((x1 – x2) ² + (y1 – y2) ²),

and find the lengths of all three sides of a given triangle with vertices at the points M (2; 3), N (-4; 6) and K (5; -1):

| MN | = √ ((2 – (-4)) ² + (3 – 6) ²) = √ ((2 + 4) ² + (3 – 6) ²) = √ (6² + 3²) = √ (36 + 9 ) = √45;

| NK | = √ ((- 4 – 5) ² + (6 – (-1)) ²) = √ ((- 4 – 5) ² + (6 + 1) ²) = √ (9² + 7²) = √ (81 + 49) = √130.

| MK | = √ ((2 – 5) ² + (3 – (-1)) ²) = √ ((2 – 5) ² + (3 + 1) ²) = √ (3² + 4²) = √ (9 + 16 ) = √25 = 5.

Using the cosine theorem, we find the cosine of the angle M

| NK | ² = | MN | ² + | MK | ² – 2 * | MN | * | MK | * cosM.

Substituting the found values ​​| MN |, | NK | and | MK |, we get:

(√130) ² = (√45) ² + 5² – 2 * √45 * 5 * cosM;

130 = 45 + 25 – 10√45 * cosM;

130 = 70 – 10√45 * cosM;

10√45 * cosM = 70 – 130;

10√45 * cosM = -60;

cosM = -60 / 10√45;

cosM = -6 / √45;

cosM = -6/3 (√5);

cosM = -2 / √5;

cosM = -2√5 / 5.

Answer: cosM = -2√5 / 5.