Question

ABC is a triangle where A = (2,3,5), B = (-1,3,2) and C = (x,5,y). If median through A is equally inclined to the axes then the length of median is?

Answer 1

Hint: We solve the above problem by using the below formulas
1. Distance formulae for 3 plane coordinates = $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $
2. To find out the coordinates of a median (x,y,z) of two points (x1,y1,z1) and (x2,y2,z2)
$\begin{gathered}
  x = \dfrac{{{x_1} + {x_2}}}{2} \\
  y = \dfrac{{{y_1} + {y_2}}}{2} \\
  z = \dfrac{{{z_1} + {z_2}}}{2} \\
\end{gathered} $
3. Direction cosines of a line from A(x1,y1,z1) to B(x2,y2,z2) is given by
\[\left( {{x_2} - {x_1},{\text{ }}{y_2} - {y_1},{\text{ }}{z_2} - {z_1}} \right)\]
4. Direction ratios of a line from A(x1,y1,z1) to B(x2,y2,z2) is given by
$\dfrac{{{x_2} - {x_1}}}{{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} }},\dfrac{{{y_2} - {y_1}}}{{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} }},\dfrac{{{z_2} - {z_1}}}{{\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} }}$

Complete step-by-step solution:
1. Assuming the triangle ABC, which has a point D on BC which has a median through A.

And assuming the coordinates of point D = (a,b,c)
Now, Point D is the median of the line BC. Which implies that
$\begin{gathered}
  a = \dfrac{{x + ( - 1)}}{2} \\
  a = \dfrac{{x - 1}}{2} \\
\end{gathered} $ ......(1)
$\begin{gathered}
  b = \dfrac{{5 + 3}}{2} \\
  b = \dfrac{8}{2} \\
  b = 4 \\
\end{gathered} $ ......(2)
And
$c = \dfrac{{y + 2}}{2}$ ......(3)
We have got the point coordinates of the point D
2. Now as to find out the values of x and y we need to find out the direction cosines of line AD.
Direction cosines of line AD is given by
$\begin{gathered}
  \left( {a - 2,b - 3,c - 5} \right) \\
    \\
\end{gathered} $
Substituting the values of a, b and c from Eqn (1), (2) and(3), we get
$\left( {\dfrac{{x - 1}}{2} - 2,4 - 3,\dfrac{{y + 2}}{2} - 5} \right)$
$\begin{gathered}
  \left( {\dfrac{{x - 1 - 4}}{2},1,\dfrac{{y + 2 - 10}}{2}} \right) \\
  \left( {\dfrac{{x - 5}}{2},1,\dfrac{{y - 8}}{2}} \right) \\
\end{gathered} $
 3. Since it was said that the median through point A makes equal angles with the axes. It means that the direction ratios of line AD are equal
i.e. $\dfrac{{\dfrac{{x - 5}}{2}}}{{\sqrt {{{\left( {\dfrac{{x - 5}}{2}} \right)}^2} + {1^2} + {{\left( {\dfrac{{y - 8}}{2}} \right)}^2}} }} = \dfrac{1}{{\sqrt {{{\left( {\dfrac{{x - 5}}{2}} \right)}^2} + {1^2} + {{\left( {\dfrac{{y - 8}}{2}} \right)}^2}} }} = \dfrac{{\dfrac{{y - 8}}{2}}}{{\sqrt {{{\left( {\dfrac{{x - 5}}{2}} \right)}^2} + {1^2} + {{\left( {\dfrac{{y - 8}}{2}} \right)}^2}} }}$
4. Comparing the first two parts of direction ratios, we get
$\dfrac{{\dfrac{{x - 5}}{2}}}{{\sqrt {{{\left( {\dfrac{{x - 5}}{2}} \right)}^2} + {1^2} + {{\left( {\dfrac{{y - 8}}{2}} \right)}^2}} }} = \dfrac{1}{{\sqrt {{{\left( {\dfrac{{x - 5}}{2}} \right)}^2} + {1^2} + {{\left( {\dfrac{{y - 8}}{2}} \right)}^2}} }}$
Since the denominators are same, they will get cancelled out and we will be left with
$\dfrac{{x - 5}}{2} = 1$
Multiply both sides with 2, we get
$x - 5 = 2$
Adding 5 both sides we will get
$x = 7$
The x-coordinate of point c is 7.
5. Comparing last two parts of direction ratios we get
$\dfrac{{\dfrac{{y - 8}}{2}}}{{\sqrt {{{\left( {\dfrac{{x - 5}}{2}} \right)}^2} + {1^2} + {{\left( {\dfrac{{y - 8}}{2}} \right)}^2}} }} = \dfrac{1}{{\sqrt {{{\left( {\dfrac{{x - 5}}{2}} \right)}^2} + {1^2} + {{\left( {\dfrac{{y - 8}}{2}} \right)}^2}} }}$
Since the denominators are same they will get cancelled out and we will be left with
$\dfrac{{y - 8}}{2} = 1$
Multiplying both sides with 2 we will get
$y - 8 = 2$
Adding 2 both sides we will get
$y = 10$
The z-coordinate of point C is 10
6. Now the point C has coordinates (7,5,10)
7. Putting the values of x and y in Eqn (1) and (2), we get
$\begin{gathered}
  a = \dfrac{{7 - 1}}{2} \\
  a = \dfrac{6}{2} \\
  a = 3 \\
\end{gathered} $
And
$\begin{gathered}
  c = \dfrac{{10 + 2}}{2} \\
  c = \dfrac{{12}}{2} \\
  c = 6 \\
\end{gathered} $
8. The coordinates of point D are (3,4,6)
Now, to find out the length of median AD, we need to find out the distance between the point A and D
9. So, using the distance formulae for three coordinate points
Distance between A(2,3,5) and D(3,4,6) will be
$\begin{gathered}
  \sqrt {{{\left( {3 - 2} \right)}^2} + {{\left( {4 - 3} \right)}^2} + {{\left( {6 - 5} \right)}^2}} \\
  \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} \\
  \sqrt {1 + 1 + 1} \\
  \sqrt 3 \\
\end{gathered} $
Therefore, the length of median AD is $\sqrt 3 $units.

Note:The median is a point always on the centre of line which bisects the line into two equal parts.
i.e. If I consider a line AB, and point C to be its median.

Then I can say that point C divides the line AB in two equal parts
i.e. AC = BC