The two adjacent vertices of a quadrilateral are (2,3) & (8,3). If the

Question

The two adjacent vertices of a quadrilateral are (2,3) & (8,3). If the third & forth vertices of the quadrilateral are (x,y) & (a,b), for how many pairs of non negative integer values of (x,y) & (a,b), the quadrilateral will be a parallelogram of area at least 12 sq. units and at most 30 sq. units?

A. 6

B. 12

C. 16

D. 20

E. >20

A. 6

B. 12

C. 16

D. 20

E. >20

chetan2u · Accepted Answer

This basically tells us that one side is parallel to x axis where y=3..
Now the length of side is 8-2=6..
The opposite side will also be 6 and parallel to x-axis..
For 12 the distance between opposite sides is at least 2, since 2*6=12
So the other line can be y=0,y=1,y=5,y=6,y=7 or y=8..
At y=8 area will be 6*(8-3)=30
But each line can have infinite points to satisfy this..
For example..
(1,1)(7,1) or (2,1)(8,1) or (3,1)(9,1) or (4,1)(10,1) or (1000,1)(1006,1).......
So infinite values

E

aviddd · Answer

Why do you assume the opposite side is parallel to x-axis. Question does not mention that this is a parallelogram.

Fdambro294 · Answer

The 1 Line Segment Side Given by the 2 Vertices = Length of 6 Units

Since its a Parallelogram, the Opposite Sides of the Quad. must be Equal and Parallel.

Furthermore, the Area of a Parallelogram is given by: (Base of 6) * (Perpendicular Height b/w Parallel Opposite Sides)

Since the Given Line Segment's Side = 6 Units, the Perpendicular Height from this Base to the OPPOSITE Side must = AT LEAST 2 Units ----- the Area will be at least 12 units squared.

We can start with the Non-Negative (X , Y) Coordinates of (0 , 0). The (A, B) Coordinate must be (6 , 0) in order for the Opposite Side of the Given Line Segment to also be 6 Units.

The Perpendicular Height drawn from the 2 Parallel Sides will = Distance Traveled on the Y-Axis. From Point (2, 3) to Point (0 , 0) this is a Height = 3.

Area = 6 * 3 = 18.

We can then take (X, Y) as (1, 0). (A, B) would then have to be (7, 0). 
The Opposite Sides would be Equal and the Area = 6 * 3 = 18 again.

We can continue taking Consecutive (X, Y) Coordinates up until (7, 0), where (A, B) = (13 , 0)

That's 8 Possibilities along the Horizontal Line of Y = 0.

Going along the Horizontal Line of Y = 1, starting at (X, Y) = (0 , 1) and (A , B) = (6 , 1),

we would similarly have 8 more Possibilities where the AREA of the parallelogram would = 6 * 2 = 12
and the Opposite Sides would be equal.

Going along the Horizontal Line of Y = 5, starting at (X , Y) = (0 , 5) and (A , B) = (6 , 5),

we would have 8 more Possibilities where the AREA of the parallelogram would = 6 * 2 = 12 
and the Opposite Sides would be Equal.

Going along Horizontal Line of Y = 6, starting at (X , Y) = (0 , 6) and (A , B) = (6 , 6),

we would have 8 more Possibilities where the AREA of the parallelogram would = 6 * 3 = 18
and the Opposite Sides would be Equal

Since we are already over > 20

the answer is -E-

Rishm · Answer

I used the parallelogram formula of base into height; we know that the coordinates are 2,3 and 8,3 respectively. Now, that makes base as 6 as 8-2 is 6. And height we need to plug in 2,3,4,5 ( Max being 12 and 30) so 6*2 is 12 and 6*5 is 30. Now it becomes obvious that there will be more than 20 different pairs, as 5 different points will form on both sides. Not sure if this approach is correct.

The two adjacent vertices of a quadrilateral are (2,3) & (8,3). If the

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The two adjacent vertices of a quadrilateral are (2,3) & (8,3). If the

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