Question

For ∆ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2), verify that a median of the triangle divides it into two triangles of equal areas.

Answer 1

To prove : Area of triangle ABD = Area of triangle ACD
Given ∆ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2).
Join AD. Let AD be the median that divides the triangle into two triangles.

So, D is the midpoint of BC.
Therefore, the coordinates of D are
$D\left(\frac{3+5}{2},\frac{-2+2}{2}\right)=D\left(\frac{8}{2},\frac{0}{2}\right)=D\left(4,0\right)$
Now,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ABD=\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left[4\left(-2-0\right)+3\left\{0-\left(-6\right)\right\}+4\left\{-6-\left(-2\right)\right\}\right] \\ =\frac{1}{2}\left\{4\left(-2\right)+3\left(6\right)+4\left(-4\right)\right\} \\ =\frac{1}{2}\left(-8+18-16\right) \\ =\frac{1}{2}\left(-6\right) \\ =3\mathrm{sq}.\mathrm{units}.\left[\mathrm{Area}\mathrm{cannot}\mathrm{be}-\mathrm{ve}\right] \end{gathered}$$
and
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ACD=\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{4\left(2-0\right)+5\left(0-\left(-6\right)\right)+4\left(-6-2\right)\right\} \\ =\frac{1}{2}\left\{4\left(2\right)+5\left(6\right)+4\left(-8\right)\right\} \\ =\frac{1}{2}\left(8+30-32\right) \\ =\frac{1}{2}\left(6\right) \\ =3\mathrm{sq}.\mathrm{units}. \end{gathered}$$
Therefore, area of triangle ABD = Area of triangle ACD
Hence, the median of ∆ABC divides it into two triangles of equal areas.