Ex.7.3 Q5 Coordinate Geometry Solution - NCERT Maths Class 10

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Question

You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for \(\Delta ABC\) whose vertices are \(A\)\((4, -6),\) \(B\)\((3, -2)\) and \(C\)\((5, 2)\)

 

Text Solution

 

Reasoning:

Let \(ABC\) be any triangle whose vertices are  \(A\)\((x_1, y_1)\), \(B\)\((x_2, y_2)\) and \(C\)\((x_3, y_3).\)

Area of a triangle
\[\begin{align}= \frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{{x}}_{{3}}}\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)} \right\} \end{align}\]

What is known?

The \(x\) and \(y\) co-ordinates of the vertices of the triangle.

What is unknown?

To verify that a median of a triangle divides it into two triangles of equal areas.

Steps:

From the figure,

Given,

  • Let the vertices of the triangle be \(A\)\((4, -6),\) \(B\)\((3, -2),\) and \(C\)\((5, 2).\)
  • Let \(M\) be the mid-point of side \(BC\) of \(\Delta ABC.\)

Therefore, \(AM\) is the median in \(\Delta ABC.\)

\[\begin{align}{\text{Coordinates of point}}\;{{M}} &= \left( {\frac{{3 + 5}}{2} + \frac{{ - 2 + 2}}{2}} \right)\\\,\,\,\,\, &= (4,0)\end{align}\]

Area of a triangle
 \[\begin{align}=\frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{{x}}_{{3}}}\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)} \right\} \\...\,\rm{{Equation}}\,\left( {{1}} \right)\end{align}\]

By substituting the values of vertices, \(A, B, M\) in the Equation (1)

Area of \(\Delta ABM\)
\[\begin{align} &= \frac{1}{2}[(4)\{ ( - 2) - (0)\} + (3)\{ (0) - ( - 6)\} + (4)\{ ( - 6)\} - ( - 2)\} \\ &= \frac{1}{2}( - 8 + 18 - 16)\\& = 3\;\text{{ Square units }}\end{align}\]

By substituting the values of vertics, \(A,D,C\) in the Equation (1)

\[\begin{align}\text{Area of }\Delta \rm{ABD} &= \frac{1}{2}[(4)\{ ( - 2) - (0)\}  + (3)\{ (0) - ( - 6)\} +(4){ ( - 6)\}  - ( - 2)}  \\&= \frac{1}{2}( - 8 + 18 - 16)\\&= 3{\text{ Square units }}\end{align}\]

However, area cannot be negative. Therefore, area of \(\Delta AMC\) is \(3\, \rm{square \,units}.\)

Hence, clearly, median \(AM\) has divided \(\Delta ABC\) in two triangles of equal areas.

  
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