Question

Find the length of the median $AD$ of a $\Delta ABC$, whose vertices are $A(5,1)$, $B(1,5)$ and $C( - 3,1)$.

Answer 1

Hint: A median of a triangle is a line segment drawn from a vertex to the midpoint of the opposite side of the vertex.
Mid-point between two points having coordinates $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.
The distance between two points having coordinates $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $.

Complete step by step answer:
We are given the $\Delta ABC$, whose vertices are $A(5,1)$, $B(1,5)$ and $C( - 3,1)$. Let $D$ be the midpoint of the side $BC$ of the $\Delta ABC$. It can be clearly shown in the diagram below.

The objective is to determine the length of the median $AD$. To determine the length of $AD$, we first need to find the coordinates of both $A$ and $D$.
Since, $D$ be the midpoint of the side $BC$, and the coordinates of $B$ and $C$ are $B(1,5)$ and $C( - 3,1)$.
Use the mid-point formula: $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$ to determine the coordinates of $D$.
Let $B(1,5) = ({x_1},{y_1})$ and $C( - 3,1) = ({x_2},{y_2})$
Substitute the above values in the midpoint formula,
$D = \left( {\dfrac{{1 + ( - 3)}}{2},\dfrac{{5 + 1}}{2}} \right)$
Solving the fractions,
$D = \left( {\dfrac{{ - 2}}{2},\dfrac{6}{2}} \right)$
Reduce the above fraction terms,
$D = \left( { - 1,3} \right)$
The coordinates of $D = \left( { - 1,3} \right)$.
Now, determine the distance between $A$ and $D$using the distance formulae given by $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $.
Let $A(5,1) = ({x_1},{y_1})$ and $D( - 1,3) = ({x_2},{y_2})$
Substitute the above values in the distance formulae:
$AD = \sqrt {{{(( - 1) - 5)}^2} + {{(3 - 1)}^2}} $
$AD = \sqrt {{{( - 6)}^2} + {{(2)}^2}} $
Evaluate the squares of the given terms,
(since the square of negative terms is positive so the square of -6 will be 36)
$AD = \sqrt {36 + 4} $
$AD = \sqrt {40} $
Evaluate the square root of 40, since it is not a perfect square so its square root is,
$AD = \sqrt {2 \times 2 \times 5 \times 2} $
$AD = 2\sqrt {10} $
Hence, $AD = 2\sqrt {10} $
The length of the median $AD$ of a $\Delta ABC$, whose vertices are $A(5,1)$, $B(1,5)$and $C( - 3,1)$ is $2\sqrt {10} $ units.

Note:
A median of a triangle refers to a line segment joining a vertex of the triangle to the midpoint of the opposite side, thus bisecting that side. For any triangle, there are exactly three medians, one from each vertex. The meeting point of all the three medians is called centroid.