# What's wrong with the cartesian plane

1. Oct 1, 2007

### electron

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 plz tell where am i wrong..

2. Oct 1, 2007

### Dick

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.

3. Oct 1, 2007

### arildno

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: Oct 1, 2007
4. Oct 1, 2007

### electron

i got the point...thankx for clearing my confusion...