# A(7, −3), B(5, 3), C(3, −1) are the vertices of

Question:

$A(7,-3), B(5,3), C(3,-1)$ are the vertices of a $\Delta A B C$ and $A D$ is its median. Prove that the median $A D$ divides $\Delta A B C$ into two triangles of equal areas.

Solution:

The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).

Coordinates of $D=\left(\frac{5+3}{2}, \frac{3-1}{2}\right)=(4,1)$

For the area of the triangle $A D C$, let $A\left(x_{1}, y_{1}\right)=A(7,-3), D\left(x_{2}, y_{2}\right)=D(4,1)$ and $C\left(x_{3}, y_{3}\right)=C(3,-1)$. Then

Area of $\Delta A D C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$=\frac{1}{2}[7(1+1)+4(-1+3)+3(-3-1)]$

$=\frac{1}{2}[14+8-12]=5$ sq. unit

Now, for the area of triangle $A B D$, let $A\left(x_{1}, y_{1}\right)=A(7,-3), B\left(x_{2}, y_{2}\right)=B(5,3)$ and $D\left(x_{3}, y_{3}\right)=D(4,1)$. Then

Area of $\Delta A B D=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$=\frac{1}{2}[7(3-1)+5(1+3)+4(-3-3)]$

$=\frac{1}{2}[14+20-24]=5$ sq. unit

Thus, Area $(\Delta A D C)=$ Area $(\Delta A B D)=5$ sq. units.

Hence, $A D$ divides $\Delta A B C$ into two triangles of equal areas.