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Area of the parallelogram formed by the lines y=mx, y=mx+1, y=nx and y=nx+1 equals,
(a) $ \dfrac{m+n}{{{\left( m-n \right)}^{2}}} $
(b) $ \dfrac{2}{{{\left( m-n \right)}^{2}}} $
(c) $ \dfrac{1}{\left( m+n \right)} $
(d) $ \dfrac{1}{\left| m-n \right|} $

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Answer
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Hint: We will first draw a figure of line y = mx, y = mx + 1 with a positive slope. Then, we will draw the lines y = nx and y = nx + 1 with a negative slope. We are taking opposite signs for the convenience of the figure and it is not a compulsory condition. From the figure, we will see that two triangles are formed with the y – axis. The area of the parallelogram is twice that of the two triangles formed.


Complete step-by-step answer:

The figure of the parallelogram will be as follows:

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The point of intersection of y = mx and y = nx + 1 is B, y = mx + 1 and y = nx + 1 is A and intersection of y = mx + 1 and y = nx is C. So, A, B, C and O form a parallelogram.

From the figure we can see that two triangles OAB and OAC are formed with OA as the common base.

OB = CA and AB = OC, since they are opposite sides of a parallelogram.

Thus, we can say that triangle OAB and triangle OAC are congruent triangles.

Therefore, the area of the parallelogram is twice the area of one of the triangles.

Area of a triangle is half the product of base and height of a triangle.

the distance of B from y – axis will be the height of the triangle OAB

We will now find point B.

To find point B, we need to solve y = mx and y = nx + 1

 $ \Rightarrow $ mx = nx + 1

 $ \Rightarrow $ (m – n)x = 1

 $ \Rightarrow $ x = $ \dfrac{1}{\left( m-n \right)} $

Thus, the height of triangle OAB is $ \dfrac{1}{\left( m-n \right)} $ .

Now, we will find the length of the base OA.

A lies on the y – axis, thus x – coordinate of A is 0.

Therefore, substitute x = 0 in y = mx + 1.

 $ \Rightarrow $ y = m(0) + 1

 $ \Rightarrow $ y = 1

Thus, the length of OA is 1 unit.

 $ \Rightarrow $ Area of parallelogram = 2(Area of triangle OAB)

 $ \Rightarrow $ Area of parallelogram = $ 2\times \dfrac{1}{2}\times 1\times \dfrac{1}{\left( m-n \right)} $

 $ \Rightarrow $ Area of parallelogram = $ \dfrac{1}{\left| m-n \right|} $

The modulus is applied as the area cannot be negative.

So, the correct answer is “Option D”.


Note: Another method to solve this will be through option verification method. Students can take any arbitrary form m and n and then find the area of parallelogram with real numbers. While taking values of m and n, students are advised to check each option and make sure each option is yielding different values.