If A (4,-6) , B(3,-2) and C(5,2) are the vertices of a ∆ABC and AD is it's median. Probe that the median AD divides ∆ABC into two triangles of equal areas.
Answers
Clearly, D is the midpoint of BC.
Therefore,
The coordinates of D are : ( 3 + 5 /2 , -2 + 2 / 2 ) = ( 4 , 0 ).
For ∆ ABD , we have
A ( X1 = 4 , Y1 = -6 ) , B ( X2 = 3 , Y2 = -2 ) and D ( X3 = 4 , Y3 = 0 ).
Ar ( ∆ ABD ) = 1/2 × [ X1 ( Y2 - Y3 ) + X2 ( Y3 - Y1 ) + X3 ( Y1 - Y2 ) ]
=> 1/2 [ 4 × ( -2 - 0 ) + 3 × ( 0 + 6 ) + 4 × ( -6 + 2 ) ]
=> 1/2 × ( -8 + 18 - 16 ) = 1/1 × | -6 |
=> 1/2 × 6 = 3 sq units.
For ∆ ADC , we have
( X1 = 4 , Y1 = -6 ) , ( X2 = 4 , Y2 = 0 ) and ( X3 = 5 , Y3 = 2 ).
Therefore,
Ar ( ∆ADC ) = 1/2 | X1 ( Y2 - Y3 ) + X2 ( Y3 - Y1 ) + X3 ( Y1 - Y2 ) .|
=> 1/2 × | 4 ( 0 - 2 ) + 4 ( 2 + 6 ) + 5 ( -6 - 0 ) |
=> 1/2 × | -8 + 32 - 30 |
=> 1/2 × | -6 |
=> 3 sq units.
Therefore,
Ar ( ABD ) = Ar ( ∆ADC )
Hence,
Meadian AD divides ∆ABC into two triangles of equal areas.
Step-by-step explanation:
It is given the A ( 4 , - 6 ) , B ( 3 , -2) and C ( 5 , 2 ) are the vertices of a ∆ABC and AD is it's median,
Clearly, D is the midpoint of BC.
Therefore,
The coordinates of D are : ( 3 + 5 /2 , -2 + 2 / 2 ) = ( 4 , 0 ).
For ∆ ABD , we have
A ( X1 = 4 , Y1 = -6 ) , B ( X2 = 3 , Y2 = -2 ) and D ( X3 = 4 , Y3 = 0 ).
Ar ( ∆ ABD ) = 1/2 × [ X1 ( Y2 - Y3 ) + X2 ( Y3 - Y1 ) + X3 ( Y1 - Y2 ) ]
=> 1/2 [ 4 × ( -2 - 0 ) + 3 × ( 0 + 6 ) + 4 × ( -6 + 2 ) ]
=> 1/2 × ( -8 + 18 - 16 ) = 1/1 × | -6 |
=> 1/2 × 6 = 3 sq units.
For ∆ ADC , we have
( X1 = 4 , Y1 = -6 ) , ( X2 = 4 , Y2 = 0 ) and ( X3 = 5 , Y3 = 2 ).
Therefore,
Ar ( ∆ADC ) = 1/2 | X1 ( Y2 - Y3 ) + X2 ( Y3 - Y1 ) + X3 ( Y1 - Y2 ) .|
=> 1/2 × | 4 ( 0 - 2 ) + 4 ( 2 + 6 ) + 5 ( -6 - 0 ) |
=> 1/2 × | -8 + 32 - 30 |
=> 1/2 × | -6 |
=> 3 sq units,
Therefore,
Ar ( ABD ) = Ar ( ∆ADC )
Hence,
Median AD divides ∆ABC into two triangles of equal areas.