Question

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

Answer 1

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (0,-1); B (2, 1) and C (0, 3)

So we should find the mid-points of the sides of the triangle.

In general to find the mid-point P(x,y) of two points `A(x_1, y_1)` and `B(x_2, y_2)` we use section formula as,

`P(x,y) = ((x_1 + x_2)/2, (y_1 + y_2)/2)`

Therefore mid-point P of side AB can be written as,

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

Now equate the individual terms to get,

x = 1

y = 0

So co-ordinates of P is (1, 0)

Similarly mid-point Q of side BC can be written as

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

Now equate the individual terms to get,

x = 1

y = 2

So co-ordinates of Q is (1, 2)

Similarly mid-point R of side AC can be written as,

`R(x,y) = ((0 + 0)/2,(3- 1)/2)`

Now equate the individual terms to get,

x =1

y= 2

So co-ordinates of Q is (1, 2)

Similarly mid-point R of side AC can be written as,

`R(x,y) = ((0 + 0)/2, (3-1)/2)`

Now equate the individual terms to get,

x = 1

y = 1

So co-ordinates of R is (0, 1)

Therefore length of median from A to the side BC is,

`AQ= sqrt((0 - 1)^2 + (-1-2)^2)`

`= sqrt(1 + 9)`

`= sqrt(10)`

Similarly length of median from B to the side AC is,

`BR = sqrt((2 - 0)^2 +(1-1)^2)`

`= sqrt4`

= 2

Similarly length of median from C to the side AB is

`CP = sqrt((0 - 1)^2 +(3 - 0)^2)`

`= sqrt(1 + 9)`

`=sqrt10`