The vertices of a triangle are …

AD is the median of {tex}\triangle{/tex}ABC from vertex A
D(x, y) = {tex}\left( \frac { 3 + 5 } { 2 } , \frac { - 2 + 2 } { 2 } \right){/tex}= (4, 0)

Area of {tex}\Delta{/tex}ADB ={tex}\frac { 1 } { 2 } \times {/tex}(4 (0 + 2) + 4 (-2 + 6) + 3 (- 6 - 0))
= {tex}\frac { 1 } { 2 } \times {/tex}(8 + 16 + -18)
= {tex}\frac { 1 } { 2 } \times {/tex}6 = 3 square units.........(i)
Area of {tex}\Delta{/tex}ACD
= {tex}\frac { 1 } { 2 } \times{/tex} ( 4(0 - 2) + 4(2 + 6) + 5 (- 6 - 0))
= {tex}\frac { 1 } { 2 } \times{/tex}(-8 + 32 - 30)
= {tex}\frac { 1 } { 2 } \times{/tex}-6 = -3
Since area can not the negative
Area of {tex}\Delta{/tex}ACD = 3 square units .........(ii)
From (i) and (ii) Area {tex}\Delta{/tex}ADB = Area {tex}\Delta{/tex}ACD, it is verified that median of ∆ABC divides it into two triangles of equal areas.