Question

Find the lengths of the medians of a ABC having vertices at A(0, -1), B (2,1) and C (0,3).

Answer 1

Explanation:

Let ABC be a triangle having vertices A(0,-1), B(2,1) and C(0,3).

Also, let AM, BN and CP be medians of △ABC and coordinates of M, N and P be (x, y), (x', y') and (x", y").

Then,

Using mid point formula,

M(x, y) = ((2+0)/2, (1+3)/2)

= (1, 2)

N(x',y') = ((0+0)/2, (3-1)/2)

= (0, 1)

and, P(x", y") = ((2+0)/2, (1-1)/2)

= (1, 0)

Now,

Using distance formula,

AM= √(1-0)²+(2+1)²

= √(1+9)

= √10 units

BN= √(0-2)²+(1-1)²

= √(4+0)

= √4

= 2 units

and, CP= √(1-0)²+(0-3)²

= √(1+9)

= √10 units

Hence,

The lengths of medians of a △ABC are √10 units, 2 units and √10 units.