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I had a question with respect to the slopes of perpendicular lines. This is a very basic question but i am slightly puzzled because I cant seem to reason it out.
The generally known formula is that if we have two lines L1 and L2 which are perpendicular and have the the slopes M1 and M2 respectively then "M1*M2= -1".
Now, lets say L1 is parallel to Y axis ( they have the same X coordinates) and L2 is parallel to X axis ( they have the same Y coordinate).These two lines intersect at point P. The formula to find the slope of a line is \(\frac{y2-y1}{x2-x1}\)
In this case, x2-x1 for L1 will be 0 and thus the slope will be undefined. However the slope M2 will be 0.
How do we apply the M1*M2 rule here as it will be ( undefined * 0). I want to understand what am i missing in this basic concept.
Regards, Shradha
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What you've described is actually the X-Y plane moved . And slope of line parallel to X axis is tan 0 and other is tan 90, which is undefined. And between these any two perpendicular lines will have slope multiplied as -1. Also slope of a line is measured with reference to X and Y axis.
What you've described is actually the X-Y plane moved . And slope of line parallel to X axis is tan 0 and other is tan 90, which is undefined. And between these any two perpendicular lines will have slope multiplied as -1. Also slope of a line is measured with reference to X and Y axis.
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Hi Senthil,
I am still unclear. I still dont get how the the slope of these two lines that i mentioned ( undefined and 0) when multiplied give a product equivalent to -1.
I am still unclear. I still dont get how the the slope of these two lines that i mentioned ( undefined and 0) when multiplied give a product equivalent to -1.
Regards, Shradha
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Ok, Let me try again. The slope is a property that exists on the plane, something like boundary conditions. At tan 90 , which is undefined, they dont apply.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.