What's wrong with the cartesian plane

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    Cartesian Plane
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The discussion addresses the confusion surrounding the slopes of the x-axis and y-axis in the Cartesian plane. The slope of the x-axis is 0, while the slope of the y-axis is undefined, leading to the misconception that their product should equal -1 for perpendicular lines. It clarifies that the product of the slopes is actually undefined, as the y-axis is vertical. The general condition for perpendicularity is explained using line equations, confirming that the specific case of the axes does not meet the criteria for the slope-product formula. The conclusion emphasizes that while the axes are indeed perpendicular, the slope concept requires careful consideration of definitions.
electron
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hi guyz,

im in skul and came across a small problem while solving some geometry questions

incase of cartesian plane..
slope of x-axis is tan(0)=0
and slope of y-axis is t(90)=infinite/not defined

so when we mutiply both the slopes we are suppsed to get -1
( from m(1)*m(2)= -1 ,, if m1 is 90 degrees to m2)

wat i hasd in mind is..are the x-axis and y-axis actually perpendicular..or they seem to be pependicular, i.e approaching to 90 degrees..

im sure there is some concept of limits over here...can u please tell where am i wrong..
 
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The ARE perpendicular and the slope of the y-axis is undefined. So the product of the slopes is undefined, not -1. m1*m2=-1 is true only if the two lines are perpendicular and neither of them is vertical.
 
Let us consider the general perpendicularity condition.
Given general line equations for lines passing through the origin,

Ax+By=0, Cx+Dy=0, where in each line equation, at least one of the coefficients is non-zero, these two lines are perpendicular iff AC+BD=0 (*).

In the case of non-zero coefficients, the slopes, with respect to the x-axis are:
a= -A/B and b=-C/D, respectively.

Thus, their product a*b equals (A*C)/(B*D), and inserting from (*), we get a*b=-1, under the requirement B&D different from zero.

The general condition for perpendicularity is seen to hold for the special choice of the ordinate axes as your two lines, whereas this particular choice of lines violates the specific conditions that hold for the slope-product formula.

As Dick has already told you..
 
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i got the point...thankx for clearing my confusion...
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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