Segment AD is the median of triangle ABC. Find the modulus of the vector AD, if A (-2; -1), B (3; 1), C (1; 5).

Let a triangle ABC be given with vertices A (- 2; – 1), B (3; 1) and C (1; 5). 1). Segment AD is the median of triangle ABC, that is, point D (x; y) is the middle of the segment BC, so its coordinates are determined as the half-sum of the corresponding coordinates of the ends of the segment: x = (3 + 1): 2 = 2; y = (1 + 5): 2 = 3. We get D (2; 3). 2). To determine the coordinates of the vector AD (x; y), it is necessary to subtract the coordinates of the beginning of the vector A (- 2; – 1) from the coordinates of the end of the vector D (2; 3), we get: x = 2 – (- 2) = 3; y = 3 – (- 1) = 4. The square of the modulus (length) of the vector AD is equal to the sum of the squares of its coordinates, that is, AD ^ 2 = x ^ 2 + y ^ 2; AD ^ 2 = 3 ^ 2 + 4 ^ 2; AD = 5.
Answer: the module of the AD vector is 5.