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Find the Lengths of the Medians of a δAbc Having Vertices at A(5, 1), B(1, 5), and C(-3, -1). - CBSE Class 10 - Mathematics

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Question

Find the lengths of the medians of a ΔABC having vertices at A(5, 1), B(1, 5), and C(-3, -1).

Solution

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (5, 1); B (1, 5) and C (-3,-1).

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) = ((5+ 1)/2, (1+5)/2)`

Now equate the individual terms to get,

x = 3

y = 3

So co-ordinates of P is (3, 3)

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

`Q(x,y) = ((1-3)/2, (5 -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) = ((5-3)/2 , (1- 1)/2)`

Now equate the individual terms to get,

x = 1

y = 0

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

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

`AQ = sqrt((5 + 1)^2 + (1- 2)^2)`

`= sqrt(36 + 1)`

`=sqrt(37)`

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

`BR = sqrt((1 - 1)^2 + (5 - 0)^2)`

`= sqrt25`

= 5

Similarly length of median from C to the side AB is

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

`= sqrt(36 + 16)`

`= 2sqrt13`

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Solution Find the Lengths of the Medians of a δAbc Having Vertices at A(5, 1), B(1, 5), and C(-3, -1). Concept: Section Formula.
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