Question

Verify that median divides a triangle into two triangles of equal areas for $ \Delta ABC $ whose vertices are $ A\left( 4,-6 \right) $ , $ B\left( 3,-2 \right) $ , $ C\left( 5,2 \right) $ .

Answer 1

Hint: We first take the vertices of $ \Delta ABC $ and try to find the median AD where D is the midpoint of BC. Then we take the midpoint theorem of \[\left( \dfrac{a+c}{2},\dfrac{b+d}{2} \right)\] for finding D. after that we find the area of the divided triangles to find the equality.

Complete step by step solution:
We have $ \Delta ABC $ whose vertices are
 $ A\left( 4,-6 \right) $ , $ B\left( 3,-2 \right) $ , $ C\left( 5,2 \right) $ .
We know that for given coordinates of vertices of
 $ \left( a,b \right);\left( c,d \right);\left( e,f \right) $ , the area of the triangle made out of those vertices will be
 $ \dfrac{1}{2}\left| a\left( d-f \right)+c\left( f-b \right)+e\left( b-d \right) \right| $ square unit.
We assume that for $ \Delta ABC $ , the median is AD where D is the midpoint of BC.
We know that for given coordinates of vertices of $ \left( a,b \right);\left( c,d \right) $ , the midpoint will be \[\left( \dfrac{a+c}{2},\dfrac{b+d}{2} \right)\].
We find the coordinates of D. So, following the mid-point theorem for $ B\left( 3,-2 \right) $ , $ C\left( 5,2 \right) $ , we get
\[D\equiv \left( \dfrac{3+5}{2},\dfrac{-2+2}{2} \right)\equiv \left( 4,0 \right)\].

Now we find the areas for $ \Delta ABD $ and $ \Delta ACD $ .
For $ \Delta ABD $ , the vertices are $ A\left( 4,-6 \right) $ , $ B\left( 3,-2 \right) $ , $ D\left( 4,0 \right) $ .
Therefore, the area is $ \dfrac{1}{2}\left| 4\left( -2 \right)+3\left( 0+6 \right)+4\left( -6+2 \right) \right|=\dfrac{6}{2}=3 $ square units.
For $ \Delta ACD $ , the vertices are $ A\left( 4,-6 \right) $ , $ C\left( 5,2 \right) $ , $ D\left( 4,0 \right) $ .
Therefore, the area is $ \dfrac{1}{2}\left| 4\left( 2 \right)+5\left( 0+6 \right)+4\left( -6-2 \right) \right|=\dfrac{6}{2}=3 $ square units.
We can see that \[ar\left( \Delta ABD \right)=ar\left( \Delta ACD \right)=3\].
Thus, Verified that the median divides a triangle into two triangles of equal areas.

Note: Instead of taking AD, we could also have taken the medians like BE and CF. The final claim of getting triangles of equal areas remains the same for all medians.