# 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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