Question

A(7, −3), B(5, 3), C(3, −1) are the vertices of a $∆ABC$ and AD is its median. Prove that the median
AD divides $∆ABC$ into two triangles of equal areas.

Answer 1

The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).
$\mathrm{Coordinates}\mathrm{of}D=\left(\frac{5+3}{2},\frac{3-1}{2}\right)=\left(4,1\right)$
For the area of the triangle ADC, let $A\left(x_1,y_1\right)=A\left(7,-3\right),D\left(x_2,y_2\right)=D\left(4,1\right)\mathrm{and}C\left(x_3,y_3\right)=C\left(3,-1\right)$. Then
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ADC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\left[7\left(1+1\right)+4\left(-1+3\right)+3\left(-3-1\right)\right] \\ =\frac{1}{2}\left[14+8-12\right]=5\mathrm{sq}.\mathrm{unit} \end{gathered}$$
Now, for the area of triangle ABD, let $A\left(x_1,y_1\right)=A\left(7,-3\right),B\left(x_2,y_2\right)=B\left(5,3\right)\mathrm{and}D\left(x_3,y_3\right)=D\left(4,1\right)$. Then
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ABD=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\left[7\left(3-1\right)+5\left(1+3\right)+4\left(-3-3\right)\right] \\ =\frac{1}{2}\left[14+20-24\right]=5\mathrm{sq}.\mathrm{unit} \end{gathered}$$
Thus, $\mathrm{Area}\left(∆ADC\right)=\mathrm{Area}\left(∆ABD\right)=5\mathrm{sq}.\mathrm{units}$.
Hence, AD divides $∆ABC$ into two triangles of equal areas.