Question

5. Using the section formula, show that the points A(1,0), B(5,3), C(2,7) and
D(-2, 4) are the vertices of a parallelogram taken in order.

Answer 1

Given : A(1,0) ,B(5,3),C(2,7),D(-2,4)

Diagonals AC and BD bisect each other. so the midpoint of AC and midpoint of BD are at the same point.

By using Section formula = (m1x₂+m2x₁)/(m1+m2),(m1y₂+m2y₁)/(m1+m2)

The midpoint of the diagonals AC is in the ratio 1:1(m1 : m2)
Here, x1= 1, x2= 2, y1= 0,y2= 7
= [(1x2) + (1x1)] /(1+1) , [(1x7) + (1x0)] /(1+1)
= (2 + 1)/2 , (7 + 0)/2
= 3/2 , 7/2 ………….. (1)

The midpoint of the diagonals BD is in the ratio 1:1(m1:m2)
Here, x1= 5, x2=- 2, y1= 3,y2= 4
= [(1 x(-2)) + (1 x 5)]/(1+1) , [(1x4) + (1x3)]/(1+1)
= (-2 + 5)/2 , (4 + 3)/2
= 3/2 , 7/2 ………………….. (2)

Hence, the two diagonals AC & BD are intersecting at the same point. So the given vertex forms a parallelogram.

HOPE THIS WILL HELP YOU...

Answer 2

Solution :

Let A( 1 ,0 ) , B( 5,3 ) , C( 2 ,7 ) and D(-2 , 4 )

are vertices of a quadrilateral .

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We know that , Diagonals bisect each

other in a parallelogram .

midpoint of  the segment joining ( x1 ,y1 )

( x2 , y2 ) is

[ ( x1+x2 )/2 , ( y1 + y2 )/2 ]

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i ) mid point of diagonal A( 1 , 0 ) and C( 2 , 7 )

= [ ( 1 + 2 )/2 , ( 0 + 7 )/2 ]

= ( 3/2 , 7/2 ) --------( 1 )

ii ) mid point of diagonal B( 5 , 3 ) and D( -2 , 4 )

= [ ( 5 - 2)/2 , ( 3 + 4 )/2 ]

= ( 3/2 , 7/2 ) ------- ( 2 )

∴ ( 1 ) = ( 2 )

ABCD is a parallelogram.

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