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 $\left({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}}\right)$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 $\left({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}}\right)$
Midpoint of the line joining A(-1,3) and B(1,-1) , D= $\left({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}}\right)$
$\Rightarrow D=\left({\displaystyle \frac{-1+1}{2}},{\displaystyle \frac{3-1}{2}}\right)=\left({\displaystyle \frac{0}{2}},{\displaystyle \frac{2}{2}}\right)=(0,1)$
Midpoint of the line joining B(1,-1) and C(5,1) , E= $\left({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}}\right)$
$\Rightarrow E=\left({\displaystyle \frac{1+5}{2}},{\displaystyle \frac{-1+1}{2}}\right)=\left({\displaystyle \frac{6}{2}},{\displaystyle \frac{0}{2}}\right)=(3,0)$
Midpoint of the line joining A(-1,3) and C(5,1) , F= $\left({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}}\right)$
$\Rightarrow F=\left({\displaystyle \frac{-1+5}{2}},{\displaystyle \frac{3+1}{2}}\right)=\left({\displaystyle \frac{4}{2}},{\displaystyle \frac{4}{2}}\right)=(2,2)$
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=$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$
  $$\begin{gathered} =\sqrt{{\left(0-5\right)}^2+{\left(1-1\right)}^2} \\ =\sqrt{25+0} \\ =5units \end{gathered}$$
Distance between B(1,-1) and F(2,2)
Length of BF=$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$
 $$\begin{gathered} =\sqrt{{\left(2-1\right)}^2+{\left(2+1\right)}^2} \\ =\sqrt{1+9} \\ =\sqrt{10}units \end{gathered}$$
Length of AE=$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$
     $$\begin{gathered} =\sqrt{{\left(3+1\right)}^2+{\left(0-3\right)}^2} \\ =\sqrt{16+9}=\sqrt{25} \\ =5units \end{gathered}$$

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.