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?

Answer 1

Answer:

Infinitely many pairs

Step-by-step explanation:

The area of the parallelogram formed is given by

$bh=6|y-3|$

since the base is 6 units and the height is $|y-3|$.

Now, we want the area to be at least 12 sq.un. and at most 30 sq.un. Which leads us to the following inequality

$12\leq 6|y-3| \leq 30$.

Dividing both sides by 6, we get  $2\leq |y-3| \leq 5$.

Now, the solution to the inequality can be written as the intersection of the solutions of $2\leq |y-3|$ and $|y-3|\leq 5$.

Solving the first inquality, we get,

$2\leq (y-3)\text{ or } (y-3)\leq -2$

now, the solution for the first inequality is

$5\leq y \text{ or } y\leq 1$

Solving the second inequality, we get,

$-5\leq y-3 \leq 5$

now, the solution for the second inequality is

$-2 \leq y\leq 8$

Taking the intersection of both solutions, we have,

$y\in [-2,1]\cup [5,8]$

So y can have the following values: 0,1,5,6,7,8

On the other hand, x can have infinitely many values, namely 0,1,2,3,4,5,...

Therefore there are infinitely many pairs of nonnegative integer values of (x,y).