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You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for $∆ABC$ whose vertices are $A (4, -6), B (3, -2)$ and $C (5, 2)$.
Given:
Vertices of a triangle are $A (4, -6), B (3, -2)$ and $C (5, 2)$.
To do:
We have to prove that the median of a triangle divides it into two triangles of equal areas.
Solution:
Let $A (4, -6), B (3, -2)$ and $C (5, 2)$ be the vertices of a triangle $ABC$.
Let $D$ be the mid-point of the side $BC$ of $\triangle ABC$.
This implies,
Coordinates of point $D =$ Midpoint of BC $= (\frac{3+5}{2}, \frac{-2+2}{2})$
$=(\frac{8}{2}, 0)$
$=(4, 0)$
$AD$ is the median of $\triangle ABC$.
Area of $\triangle \mathrm{ABD}=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
$=\frac{1}{2}[4 \times(-2-0)+3(0+6)+4(-6+2)]$
$=\frac{1}{2}[-8+18-16]$
$=\frac{-6}{2}$
$=3$ sq units
Area of $\triangle \mathrm{ADC}=\frac{1}{2}[4(0-2)+4(2+6)+5\times(-6-0)]$
$=\frac{1}{2}[-8+32-30]$
$=\frac{-6}{2}$
$=3$ sq units
Area of $\triangle \mathrm{ABD}=$Area of $\triangle \mathrm{ADC}$
Therefore, the median of a triangle divides it into two triangles of equal areas.