Area Of The Region Bounded By The Curves - Concepts & Shortcuts



  • Statutory warning : It's always best to sketch the curves than mugging up the formulae.

    Some useful formulae:

    Area bounded by the curves |ax +/- m | = p and |by +/- n| = q is 4pq/ab sq units
    Area bounded by |ax +/- m| + |by +/- n| = k is 2k^2/(ab)
    Area bounded by |ax + by| = k and |ax - by| = k is 2k^2/ab
    Area bounded by |ax + by| + |ax - by| = k is k^2/(ab)

    We will see in detail how we got these fancy formulas and how easy it is to derive them whenever you need it!

    Area of a square with diagonal d = d^2/2
    Area of a rhombus with diagonal d1 and d2 = (d1 x d2)/2
    Area of parallelogram = base x height

    Some important graphs

    |x| = k (say k = 2) will give 2 lines parallel to y axis. One of which represents x = 2 and the other x = - 2
    Similarly |y| = k will give 2 lines parallel to x axis. One of which represents y = 2 and the other y = - 2

    So |x| = 2 AND |y| = 2 we will give us a square with side = 2k as shown below.

    0_1508386814167_6913035a-962d-4ca2-9385-83e1346dd52f-image.png

    Now what about |x + 3| = 2 AND |y + 1| = 2 ?

    |x + 3| = 2 => x = -1 or x = -5 (4 units)
    |y + 1| = 2 => y = 1 or y = -3 ( 4 units)
    So this should also give a square of side 4 units

    Graph is as below

    0_1508387994575_2c8cd950-4682-49c4-989a-5c93e03e590e-image.png

    One interesting thing here is even if we plot |x + 6| = 2 and |y + 8| = 2
    |x + 6| = 2 => x = -8 or x = -4 (4 units)
    |y + 8| = 2 => y = -6 or y = -10 (4 units)
    It will still give a square of side 4 units.

    Or say if we plot |x - 3| = 2 and |y + 4| = 2
    |x - 3| = 2 => x = 1 or x = 5 (4 units)
    |y + 8| = 2 => y = -6 or y = -10 (4 units)
    again, a square of side 4 units.

    So all the values of |x +/- a| = k and |y +/- b| = k will yield a square of side 2k units. (definitely in different coordinates depending on a and b)

    What if the value of k is different. Like |x + 3| = 1 and |y - 1| = 2 ?
    |x + 3| = 1 => x = -2 or x = -4 (2 units)
    |y - 1| = 2 => y = 3 or y = -1 (4 units)
    A rectangle with sides 2 and 4.
    We get a rectangle instead of square. That's it. Graph will be like

    0_1508393883586_258b6222-02ed-4899-8be1-b5546bacf4f9-image.png

    We can say in general |ax +/- m | = p and |by +/- n| = q will plot a rectangle with side 2p/a and 2q/b.
    Hence area = 4pq/ab.

    Now we will see another important type.

    |x| + |y| = 2 will give a square with diagonal = 2k as shown below.

    0_1508387365681_3adc7b85-4e7a-4282-a39b-bf9fb9d1edb2-image.png

    Here also if you plot any |x +/- a| + |y +/- b| = 2 it will still yield a square with diagonal = 2k.

    So in general, |x +/- a| + |y +/- b| = k will yield a square of diagonal 2k.

    What about |3x| + |4y| = 12 ?

    Graph is as below

    0_1508389988480_cc76a9bc-b680-40a0-8fd8-6eb462b57cad-image.png

    here it is a rhombus with diagonals 6 and 8 (which is nothing but 12 x 2/3 and 12 x 2/4, where 3 and 4 are our coefficients). Area here is (2 x 12^2)/(3x4)

    Here also if you try something like |3x + 6| + |4y - 4| = 12, it will plot the same shape (a rhombus with diagonals as 6 and 8 and the only change will be the position of the rhombus in the xy plane which won't alter the area)

    So in general, Area of |ax +/- m| + |by +/- n| = k is 2k^2/ab

    We saw the graph of |x| + |y| = k, what about the graph of |x| - |y| = k ?

    For example graph of |x| - |y| = 3 is as below

    0_1508391204556_824ab58a-e667-45ed-964b-74ecefccfaf7-image.png

    means it won't bound any region in the xy plane.

    what about |x - y| = k type ?

    For example, graph of |x - y| = 3 looks like

    0_1508391353751_73068cb7-a496-49ed-93ab-261f2f854458-image.png

    so this one also won't bound any region in the xy plane (good, lesser formulas!)

    |x + y| = k case is also same. for example |x + y| = 3 won't bound any region in the xy plane. Graph is as below

    0_1508391494263_7658fccb-573b-486d-a931-c364ce7cb6fe-image.png

    What if we combine both ? i.e |x + y| = 3 and |x - y| = 3. Can you guess from the above graphs how it would turn out ?

    Bounded region is as below

    0_1508391651605_c138fab1-df5a-4ffb-8bf0-58497624b736-image.png

    A square with diagonal as 6.

    For example, the graph of |3x + 2y| = 6 and |3x - 2y| = 6 is shown as below

    0_1508392290593_5bc8640f-1d67-4d02-8695-450cd5353f13-image.png

    So in general, if we plot |ax + by| = k and |ax - by| = k, we will get a rhombus with diagonals 2k/a and 2k/b.

    Okay. so now we can solve the graph for |x + y| = 4 and |x - y| = 4. But what about |x + y| + |x - y| = 4 ?

    Graph is as below

    0_1508322923396_e0797571-63d6-4a27-bb9e-b988abf8c487-image.png

    We can see that the graph will give a square of side = k
    so area = k^2 = 16

    what would be the graph of |2x + y| + |2x - y| = 8 ?

    Graph will be like

    0_1508324907775_2b6ffa31-1510-4c9d-8843-ecd44cf6a984-image.png

    Rectangle with side 8 and 4. Area = 8 x 4 = 32

    If you see this, the sides are nothing but 8/2 and 8/1. Where 2 and 1 are nothing by coefficients of x and y

    So we can say that the area covered by the graph |ax + by| + |ax - by| = k is k/a * k/b = k^2/ab

    How to plot the graph of a line, say 3x + 2y = 6 ?

    when x = 0, y = 3
    when y = 0, x = 2

    Graph is as below

    0_1508393666893_dde4d6ac-f84d-4e10-a07d-4c265caadac3-image.png

    x^2 + y^2 = r^2 is the equation of a circle with radius = r and center at origin.

    For example, x^2 + y^2 = 9 will plot a circle as below

    0_1508394946852_7c6dab68-a2d5-4a59-a605-c85b4630ead2-image.png

    We have compiled some questions from this topic for your practice and can be referred at Question Bank - Area Of The Region Bounded By Curves

    Share the formulas/concepts which we missed out and point out errors (if any).


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