VERIFY THAT THE GIVEN POINTS ARE VERTICES OF A PARALLELOGRAM

(i)  Find the length of all sides.

(ii) In a parallelogram, the length of opposite sides will be equal.

Question 1 :

A(4, 6), B(7, 7) C(10, 10) and D (7, 9)

Solution :

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x2 - x1)2 + (y2 - y1)2

Length of AB :

Here x1 = 4, y1 = 6, x2 = 7  and  y2 = 7

=  √(7-4)2 + (7-6)2

=  √32+12

=  √10

Length of BC :

Here x1 = 7, y1 = 7, x2 = 10  and  y2 = 10

 =  √(10-7)2 + (10-7)2

  =  √32+32

  =  √(9+9)

  =  √18 units

Length of CD :

Here x1 = 10, y1 = 10, x2 = 7  and  y2 = 9

 =  √(7-10)2 + (9-10)2

  =  √(-3)² + (-1)²

  =  √9 + 1

  =  √10 units

Length of DA :

Here x1 = 7, y1 = 9, x2 = 4 and  y2 = 6

 = √(4-7)² + (6-9)²

  =  √(-3)² + (-3)²

  =  √9 + 9

  =  √18 units

Length of opposite sides are equal. So, the given vertices will form a parallelogram.

Question 2 :

Examine whether the given points 

A(3, -5), B(-5, -4), C (7, 10) and D (15, 9)

forms a parallelogram.

Solution :

Length of AB :

Here x1 = 3, y1 = -5, x2 = -5 and  y2 = -4

=  √(-5-3)2 + (-4-(-5))2

=  √(-8)2+(-4+5)2

=  √(64+1)

=  √65 units

Length of BC :

Here x1 = -5, y1 = -4, x2 = 7 and  y2 = 10

=  √(7-(-5))2 + (10-(-4))2

=   √(7+5)2 + (10+4)2

=   √122 + 142

=   √144 + 196

=  √340 units

Length of CD :

Here x1 = 7, y1 = 10, x2 = 15 and  y2 = 9

=  √(15-7)2 + (9-10)2

=  √82 + (-1)²

=  √64 + 1

=  √65 units

Length of DA :

Here x1 = 15, y1 = 9, x2 = 3 and  y2 = -5

=  √(3-15)² + (-5-9)²

=  √(-12)² + (-14)²

=  √144 + 196

=  √340 units

Length of opposite sides are equal. So, the given vertices will form a parallelogram.

Question 3 :

Examine whether the given points

A (-4, -3) and B (3, 1) and C (3, 6) and D (-4, 2)

forms a parallelogram.

Solution :

Length of AB :

Here x1 = -4, y1 = -3, x2 = 3 and  y2 = 1

=  √(3-(-4))2+(1-(-3))2

=   √(3+4)2+(1+3)2

=  √72+42

=  √49+16

=  √65 units

Length of BC :

Here x1 = 3, y1 = 1, x2 = 3 and  y2 = 6

=  √(3-3)2 + (6-1)2

=   √5²

=  5 units

Length of CD :

Here x1 = 3, y1 = 6, x2 = -4 and  y2 = 2

 √(-4-3)2 + (2-6)2

=   √(-7)2 + (-4)2

=  √(49+16)

=  √65 units

Length of DA :

Here x1 = -4, y1 = 2, x2 = -4 and  y2 = -3

=  √(-4-(-4))² + (-3-2)²

=   √(-4+4)² + (-5)²

=  √25

=  5 units

Length of opposite sides are equal. So, the given vertices will form a parallelogram.

Question 4 :

Examine whether the given points 

A(8, 4), B(1, 3), C(3, -1) and D (4, 6)

forms a parallelogram.

Solution :

Length of AB :

Here x1 = 8, y1 = 4, x2 = 1 and  y2 = 3

=  √(1-8)² + (3-4)²

=   √(-7)² + (-1)²

=  √49 + 1

=  √50 units

Length of BC :

Here x1 = 1, y1 = 3, x2 = 3 and  y2 = -1

=  √(3-1)² + (-1-3)²

=   √(2)² + (-4)²

=  √4 + 16

=  √20 units

Length of CD :

Here x1 = 3, y1 = -1, x2 = 4 and  y2 = 6

=  √(4-3)² + (6-(-1))²

=  √(1)² + (6+1)²

=  √1 + 7²

=  √1 + 49

=  √50 units

Length of DA :

Here x1 = 4, y1 = 6, x2 = 8 and  y2 = 4

=  √(8-4)2 + (4-6)2

=  √42 + (-2)2

 √16 + 4

=  √20 units

Length of opposite sides are equal. So the given vertices forms a parallelogram

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