Consider two straight lines X'OX and YOY' and these two lines are perpendicular to each other. Draw them on a plain graph paper.
Theße lines divide the plane into four regions - we call them as quadrants (derived from the word quater)
Division of coordinate plane by X'OX and YOY'
The quater that falls under zone of OX & YO is said First; it has all positive ordered pair of real numbers (x, y)
The quater under X'O and OY' is Third; it has all negative pairs, that is (-x, -y)
Second quadrant has (-x, y) and falls in zone of X'O and YO; it's inverse is Fourth quadrant that is (x, -y) and falls under OX and OY'
Hypothetical Scenarios
Consider a hypothetical scenario in which the two axes X'OX and YOY' are not perpendicular to each
We mean to say, that would change the whole scenary. Let's say YOY' is a little bent towards the OX zone. Due to that there will be a projection (shadow) of YOY' on OX and X'O which we have to nullify to express the coordinates of a point... That's an interesting scenario to resolve. We call this as “oblique axes”
Let me show you an image of the scenario, so that we could proceed together for a solution :
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| Here we need to check_mate the problem of extra shadow Y times cosØ which adds up on x -axis |
Here the problem is projection YcosØ which is creating a problem. Due to this, a point (x,y) will now need to have adjusted to a smaller reading in X -axis as Ycosø would be increasing it; also we need to increase new Y reading or ordinate because Ysinø is lesser than complete Y.
Y = $\sqrt{(Ysinø)² + (YcosØ)²}$
New reading (X, Y) would be like wise...
X = x- ycosØ ; and Y=y+ ysinØ (probably; I am also not sure whether it's correct)
Let's try putting Thetha as 30°,
Then a pt (3,4) would then correspond to
X=3 - 4 × √3/2 = 3- 2√3 = 3 - 3.4 = - 0.4
Y = 4+ 4*1/2 = 4+2 = 6
Means a point (3,4) corresponds to a New point (-0.4, 6).
Again I say that I may be wrong completely. Let's have a discussion in comments
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