Given,

∆ABC and their vertices A (3, 5), B (7, 8) and C (1, – 10).

And, AD is median

So, D is mid-point of BC

Hence, the co-ordinates of D is ([7 + 1]/2, [8 – 10]/2) = (4, -1)

Now,

Slope of AD, m = y₂ – y₁/ x₂ – x₁

m = (5 + 1)/ (3 – 4) = 6/-1 = -6

Thus, the equation of AD is given by

y – y₁ = m (x – x₁)

y + 1 = -6 (x – 4)

y + 1 = -6x + 24

⇒ 6x + y – 23 = 0