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Slope of the perpendicular line on the other line 
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#1
Apr914, 02:05 AM

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Why is the slope of perpendicular line on the the other defined this way:
m = slope of a particular line m'= slope of the perpendicular line on that particular line m*m' = 1 OR m' = 1/m Thanks 


#2
Apr914, 04:25 AM

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If you have a line y=mx+c, and another line y'=m'x+c', what does m' have to be for y' to be perpendicular to y.
Draw a few lines and see. i.e. if m=m', then the two lines are parallel. 


#3
Apr914, 07:59 AM

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#4
Apr914, 08:03 AM

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Slope of the perpendicular line on the other line
One way to look at it is this: if y= mx+ b is the equation of a line, then the slope, m, is the tangent of the angle, [itex]\theta[/itex], the line makes with the xaxis. If two lines are perpendicular then they form a right triangle with the xaxis as hypotenuse. The angle one of the lines makes with the x axis, say [itex]\theta[/itex], will be acute, the other, [itex]\phi[/itex], will be obtuse.
Looking over this I see I have used the wrong words. I meant that one will be less that or equal to 45 degrees, the other larger than or equal to 45 degrees. The angles inside that right triangle will be [itex]\theta[/itex] and [itex]\pi \phi[/itex] and we must have [itex]\theta+ (\pi \phi)= \pi/2[/itex] so that [itex]\phi \theta= \pi/2[/itex]. Now use [itex]tan(a+ b)= \frac{tan(a)+ tan(b)}{1 tan(a)tan(b)}[/itex] with [itex]a= \phi[/itex] and [itex]b= \theta[/itex]: [tex]tan(\phi \theta)= \frac{tan(\phi) tan(\theta)}{1+ tan(\phi)tan(\theta)}= tan\left(\frac{\pi}{2}\right)[/tex] But [itex]tan(\pi/2)[/itex] is undefined! We must have the fraction on the left undefined which means the denominator must be 0: [itex]1+ tan(\theta)tan(\phi)= 0[/itex] so that [itex]tan(\theta)tan(\phi)= 1[/itex] and that last is just "[itex]mm'= 1[/itex]". 


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