Question

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

Answer 1

Hint: In order to find the length of the medians we need to fint the midpoint of the sides of the triangle using the formula $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$and the length of the medians is found by using the distance formula $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

Complete step-by-step answer:
We are given the vertices of the triangle to be A (-1,3) , B(1,-1) and C(5,1)

Medians are the lines joining the midpoint of a side of a triangle to the opposite vertice.
So we need to find the midpoint of the three sides first
We know the midpoint of the line joining the points $\left( {{x_1},{y_1}} \right)$and $\left( {{x_2},{y_2}} \right)$ is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Midpoint of the line joining A(-1,3) and B(1,-1) , D= $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
$ \Rightarrow D = \left( {\dfrac{{ - 1 + 1}}{2},\dfrac{{3 - 1}}{2}} \right) = \left( {\dfrac{0}{2},\dfrac{2}{2}} \right) = (0,1)$
Midpoint of the line joining B(1,-1) and C(5,1) , E= $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
$ \Rightarrow E = \left( {\dfrac{{1 + 5}}{2},\dfrac{{ - 1 + 1}}{2}} \right) = \left( {\dfrac{6}{2},\dfrac{0}{2}} \right) = (3,0)$
Midpoint of the line joining A(-1,3) and C(5,1) , F= $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
$ \Rightarrow F = \left( {\dfrac{{ - 1 + 5}}{2},\dfrac{{3 + 1}}{2}} \right) = \left( {\dfrac{4}{2},\dfrac{4}{2}} \right) = (2,2)$
Now we have the medians CD ,BF and AE

To find their length we need to use the distance formula
That is the distance between two points $\left( {{x_1},{y_1}} \right)$and $\left( {{x_2},{y_2}} \right)$ is $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Therefore the length of medians is given by
Distance between C(5,1) and D(0,1)
Length of CD=$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
  $
   = \sqrt {{{\left( {0 - 5} \right)}^2} + {{\left( {1 - 1} \right)}^2}} \\
   = \sqrt {25 + 0} \\
   = 5units \\
$
Distance between B(1,-1) and F(2,2)
Length of BF=$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
 $
   = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {2 + 1} \right)}^2}} \\
   = \sqrt {1 + 9} \\
   = \sqrt {10} units \\
$
Length of AE=$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
     $
   = \sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( {0 - 3} \right)}^2}} \\
   = \sqrt {16 + 9} = \sqrt {25} \\
   = 5units \\
$

Therefore the length of the medians are 5 units ,$\sqrt {10} units $ and 5 units.

Note: The point of co incidence of the medians is known as centroid.
The point of co incidence of the altitudes is known as orthocentre.