What's wrong with the cartesian plane

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    Cartesian Plane
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Homework Help Overview

The discussion revolves around the properties of the Cartesian plane, specifically the slopes of the x-axis and y-axis and their relationship in terms of perpendicularity in geometry.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions whether the x-axis and y-axis are truly perpendicular, given the undefined slope of the y-axis and the implications of multiplying slopes. Some participants explore the general condition for perpendicularity and clarify the misunderstanding regarding the product of slopes.

Discussion Status

Participants are actively engaging in clarifying the concepts of slopes and perpendicularity. Some guidance has been provided regarding the conditions under which the product of slopes equals -1, and the original poster expresses gratitude for the clarification received.

Contextual Notes

The discussion includes considerations of limits and the definitions of slopes in the context of vertical and horizontal lines, as well as the specific conditions required for the slope-product formula to hold true.

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..
 
Last edited:
i got the point...thankx for clearing my confusion...
 

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