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

[zabeer]

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.

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

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

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.

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

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

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

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

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

A square with diagonal as 6.

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

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

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

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

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

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