A median of a triangle divides it into two triangles of equal areas

Solution:

Here, the vertices of the triangles are A(4, –6), B(3, –2) and C(5, 2).

Let D be the midpoint of BC.

$\therefore \quad$ The coordinates of the mid point D are

$\therefore \quad$ The coordinates of the mid point $D$ are

$\left\{\frac{\mathbf{3}+\mathbf{5}}{\mathbf{2}}, \frac{-\mathbf{2}+\mathbf{2}}{\mathbf{2}}\right\}$ or $(4,0)$

Since, $A D$ divides the triangle $A B C$ into two parts i.e., $\triangle A B D$ and $\triangle A C D$,



Now, $\operatorname{ar}(\Delta \mathrm{ABD})$

$=\frac{1}{2}[4\{(-2)-0\}+3(0+6)+4(-6+2)]$

$=\frac{1}{2}[(-8)+18+(-16)]=\frac{1}{2}(-6)=-3$ sq. units

$=3$ sq. units(numerically) .....(1)

$\operatorname{ar}(\Delta \mathrm{ADC})=\frac{\mathbf{1}}{\mathbf{2}}[4(0-2)+4(2+6)+5(-6-0)]$

$=\frac{\mathbf{1}}{\mathbf{2}}[-8+32-30]=\frac{\mathbf{1}}{\mathbf{2}}[-6]=-3$ sq.units

$=3$ sq.units (numerically) ....(2)

From (1) and (2)

$\operatorname{ar}(\Delta \mathrm{ABD})=\operatorname{ar}(\Delta \mathrm{ADC})$

i.e., A median divides the triangle into two triangles of equal areas.