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Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nx and y = nx + 1 equals (2001S)
- a)|m + n|/(m - n)2
- b)2/|m + n|
- c)1/(|m + n|)
- d)1/(|m - n|)
Correct answer is option 'D'. Can you explain this answer?
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Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = ...
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Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = ...
Solution:
Given lines are:
y = mx
y = mx + 1
y = nx
y = nx + 1
Let A = (0, 0), B = (1, m), C = (1, m + 1), D = (1, n), E = (1, n + 1), F = (0, n)
Now, AB = √(1 + m^2), BC = √(1 + (m + 1)^2), CD = √(1 + n^2), DE = √(1 + (n + 1)^2), EF = √(1 + n^2), FA = √(1 + m^2)
Let AF = a, AB = b, BC = c, CD = d, DE = e, EF = f
a = √(1 + m^2), b = √(1 + m^2), c = √(1 + (m + 1)^2), d = √(1 + n^2), e = √(1 + (n + 1)^2), f = √(1 + n^2)
Using the formula for the area of the parallelogram formed by the vectors AB and AF,
Area = |AB x AF| = |(1 + m^2)^(1/2) (1 + m^2)^(1/2) sinθ|
= |(1 + m^2)^(1/2) (1 + m^2)^(1/2) (n - m)/(m - n)|
= |(n - m)/(m - n)| (1 + m^2)^(1/2) (1 + n^2)^(1/2)
= |(n - m)/(m - n)|/(mn)^(1/2)
= 1/|(m - n)/(mn)^(1/2)|
= 1/|m - n|
Hence, option (D) is the correct answer.
Given lines are:
y = mx
y = mx + 1
y = nx
y = nx + 1
Let A = (0, 0), B = (1, m), C = (1, m + 1), D = (1, n), E = (1, n + 1), F = (0, n)
Now, AB = √(1 + m^2), BC = √(1 + (m + 1)^2), CD = √(1 + n^2), DE = √(1 + (n + 1)^2), EF = √(1 + n^2), FA = √(1 + m^2)
Let AF = a, AB = b, BC = c, CD = d, DE = e, EF = f
a = √(1 + m^2), b = √(1 + m^2), c = √(1 + (m + 1)^2), d = √(1 + n^2), e = √(1 + (n + 1)^2), f = √(1 + n^2)
Using the formula for the area of the parallelogram formed by the vectors AB and AF,
Area = |AB x AF| = |(1 + m^2)^(1/2) (1 + m^2)^(1/2) sinθ|
= |(1 + m^2)^(1/2) (1 + m^2)^(1/2) (n - m)/(m - n)|
= |(n - m)/(m - n)| (1 + m^2)^(1/2) (1 + n^2)^(1/2)
= |(n - m)/(m - n)|/(mn)^(1/2)
= 1/|(m - n)/(mn)^(1/2)|
= 1/|m - n|
Hence, option (D) is the correct answer.
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Question Description
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