**priyanshu kumar**

Let A(0,-1) , B(2,1) ,C(0,3) are vertices of the

Triangle.

D , E , F are midpoints of BC , CA and AB.

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The midpoint of the line segment joining

the points (x1,y1) and (x2 , y2 ) is P( x , y ).

x = ( x1 + x2 )/2 ;

y = ( y1 + y2 )/2

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Now ,

i ) mid point of B(2,1) , C(0,3) is D( x , y )

x = ( 2 + 0 )/2 = 1

y = ( 1 + 3 )/2 = 2

D = ( 1 , 2 )

Similarly ,

ii ) mid point of C( 0,3) , A(0,-1) = E( 0,1)

iii ) mid point of A(0,-1), B(2,1) = F(1,0)

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The area of the triangle formed by the

vertices ( x1,y1 ), ( x2, y2 ) , ( x3 , y3 ) is

1/2|x1 ( y2 - y3 )+x2( y3 - y1 ) +x3( y2 - y1) |

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iv ) Area of the triangle A( 0, -1), B( 2 , 1 )

C( 0 ,3 ) is

1/2|0( 1 - 3 ) + 2( 3 + 1 ) + 0 ( -1-1 ) |

= 1/2 | 8 |

= 4 sq units

v ) Area of the triangle D( 1,2 ) , E( 0 , 1 ),

and F( 1 , 0 ) is

1/2 | 1( 1 - 0 ) + 0( 0 - 2 ) + 1( 2 - 1 ) |

= 1/2 | 2 |

= 1 sq units

vi )

ratio = ( area ∆ABC )/( area ∆DEF )

= 4/1

= 4 : 1

*-Answered by priyanshu kumar* On 02 October 2019:12:32:50 PM