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Find the Lengths of the Medians of a Triangle Whose Vertices Are a (−1,3), B(1,−1) and C(5, 1). - Mathematics

Find the lengths of the medians of a triangle whose vertices are A (−1,3), B(1,−1) and C(5, 1).

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Solution

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (−1, 3); B (1,−1) and C (5, 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) = ((-1+1)/2 , (3- 1)/2)`

Now equate the individual terms to get,

x= 0

y = 1

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

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

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

Now equate the individual terms to get,

x = 3

y = 0

So co-ordinates of Q is (3, 0)

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

`R(x,y)  = ((5 - 1)/2,(1+3)/2)`

Now equate the individual terms to get,

x = 2

y = 2

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

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

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

`= sqrt(16 + 9)`

= 5

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

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

`= sqrt(1 + 9)`

`= sqrt(10)`

Similarly length of median from C to the side AB is

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

`= sqrt25`

= 5

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.3 | Q 18 | Page 29
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