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High school geometry
Course: High school geometry > Unit 6
Lesson 5: Equations of parallel & perpendicular lines- Parallel lines from equation
- Parallel lines from equation (example 2)
- Parallel lines from equation (example 3)
- Perpendicular lines from equation
- Parallel & perpendicular lines from equation
- Writing equations of perpendicular lines
- Writing equations of perpendicular lines (example 2)
- Write equations of parallel & perpendicular lines
- Proof: parallel lines have the same slope
- Proof: perpendicular lines have opposite reciprocal slopes
- Analytic geometry FAQ
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Proof: perpendicular lines have opposite reciprocal slopes
CCSS.Math:
To prove that perpendicular lines have opposite reciprocal slopes, draw the two lines and label key points. By identifying the two right triangles that are formed, one can use the definition of slope to calculate the slopes of the two lines. Finally, using the properties of similar right triangles, show that the two slopes are indeed opposite reciprocals of each other.
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Line DE is negative because it goes from D to E in the direction. He could have just gone in the direction of E to D and had line ED that doesn't need the negative symbol because its going in a positive direction! For that matter why did he make the line CB and not BC which would have needed the negative symbol? If any of the lines are changed as I mentioned then the proof would fail. So what am I not understanding?(18 votes)
If a line goes forward (to the right) in the upper right quadrant it has a positive slope, and if it goes to the lower right side it has a negative slope. This is reciprocated in the other direction (positive slope = bottom left and negative slope = top left). The ED/DE line goes to the right in the lower right quadrant, so the slope must be negative, unlike the BC/CB one which goes into the upper right quandrant.(20 votes)
Why do perpendicular lines have a negative reciprocal slope ? I'm still a bit puzzled(3 votes)
Because perpendicular lines are completely opposite of each other to get the ninety degree angle,( their number one characteristic( which I suppose you know...)). They need to differ each other in every possible way from the first one or else they might intersect, but not at the desired ninety degree angle.
Hope this helps!(7 votes)
At3:11, Sal said that we have to take DE as negative then why didn't he took BC as negative on LHS? Isn't it also dropping?(3 votes)
He is saying DE is negative while AD is positive so that the slope of AE is negative, so he is using this fact for a purpose, not just to say that it is negative for no reason. On the other hand, if you start from A, you are correct that BC is negative, but AB would also be negative, so when you talk about the slope of AC, when you divide a negative by a negative, you would get a positive.
Hope this helps(5 votes)
- Isn't the similarity postulate supposed to be 'AA', instead of 'AAA'?(3 votes)
Different geometry textbooks and resources use one or the other, but it means the same thing because if two angles of a triangle are equal, the third is also.(5 votes)
I think I finally understood how it worked when you drew the triangles. I get it now! When you look at one perpendicular line, and then you look at the second perpendicular line, it's almost as if the rise and run SWITCHED PLACES! That's why the slope of a perpendicular line is the negative reciprocal of the other perpendicular line!
Edit: That IS how it works, right?(4 votes)- Almost. That only works if the triangles formed from the perpendicular lines are congruent (i.e. having the exact same sides). For instance, if you made the triangle in the right bigger than the left (while still keeping their similarity), you would get a totally different rise and a totally different run.(0 votes)
- How do you know how properly approach to the problem? I mean if I was going to prove such things I would go in the wrong direction and stuck there. Is there some common sense of proving things like this? Thanks!(2 votes)
Sometimes you can look for clues, patterns. Examine how you approached similar problems in the past. Other times, you got to pick a path and go through with it. The first person who came up with this particular problem didn't know what direction to go in, or even if there was any direction at all. If you get stuck, try a new angle.
If you try and try and try and try to do something, like how people tried to find a fractional representation of the square root of two, and it's just not working, maybe instead of trying to find a possible answer, you could prove why it's impossible. That's what Euclid did. He did something called a proof by contradiction, where you assume the contrary thing is true at the start of the proof. It's worth reading about. Hope this is enough of an answer to get you started, because really, this is what the joy of mathematics is all about.(4 votes)
I could take the slope of L as the line AC and the slope will be negative not positive and the the slope Of M will not be negative reciprocal of L(2 votes)- what do you mean the slope of AC is negative, it is going up from left to right which indicates a positive slope. Slope of m is going down from leftto right which indicates that it is a negative slope. That does indicate that the slopes of the two lines are opposites, so then you have to show that they are alos reciprocal slopes to be perpendicular.(3 votes)
- Can we use dot products for deriving the same result?(2 votes)
Can someone please explain this to me? Because I'm not able to process this that well, therefore I am unable to understand.(2 votes)
is there an easier way to get the same answer.(2 votes)
3:11Video transcript
- [Voiceover] What I'd
like to do with this video is use some geometric
arguments to prove that the slopes of perpendicular lines are negative reciprocals of each other. And so, just to start off. We have lines L and M and
we are going to assume that they are perpendicular. So, they intersect at a right angle. We see that depicted right over here. And so, I'm gonna now
construct some other lines here to help us make our geometric argument. So, let me draw a horizontal
line that intersects this point right over here. Let's call that point A. And so, let me see if I can do that. There you go. So, that's a horizontal
line that intersects at A. And now, I'm gonna drop
some verticals in that. So, I'm gonna drop a vertical
line right over here, and I'm gonna drop a vertical
line right over here. And so, that is 90 degrees,
and that is 90 degrees. And I've constructed it that way. This top line is perpendicular horizontal. And then, I've dropped to vertical things. So, there are at 90 degree angles. And let me know setup some points. So, that I already said, that's point A. Let's call this point B. Let's call this point C. Let's call this point D. And let's call this
point E, right over here. Now, let's think about
what the slop of line L is. So, slope of L is going to be what? Well, you view line L as
the line that connects point CA, so it's the
slope of CA, you could say. This is the same thing
as slope of line CA. L is line CA. And so, to find the
slope, that's change in Y over change in X. So, our change in Y is going to be CB. So, it's gonna be the length of CB. That is our change in Y. So, it is CB over our change in X, which is the length of segment of BA, which is the length of BA right over here. So, that is BA. Now, what is the slope of line M? So, slope of M. And we could also say slope
of, we could call line M line AE. Line AE, like that. Well, if we're going to go
between point A and point E, once again, it's just change
in Y over change in X. Well, what's our change in Y going to be. Well, our change in Y. Well, we're gonna go from
this level down to this level, as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So, we might be tempted to say, well that's just gonna be
the length of segment DE. But remember, our Y is decreasing. So, we're gonna subtract
that length as we go from this Y level, to that Y level over there. And what is our change in X? So, our change in X, as we go form A to E, our change in X is going to
be the length of segment AD. So, AD. So, our slope of M is
going to be negative DE. That's going to be the
negative of this length, cause we're dropping by that much. That's our change in Y, over segment A, over segment AD. So, some of you might
already be quite inspired by what we've already written, because now we just have to established that these two triangles, triangle CBA and triangle ADE are similar, and then we're going to be able to show that these are the negative
reciprocal of each other. So, let's show that these
two triangles are similar. So, let's say that we have
this angle right over here. And let's say that angle has
measure X, just for kicks. And let's say that we have,
let me do another color for, let's say we have this
angle right over here. And let's say that the measure
that that has measure Y. Well, we know X + Y + 90 is equal to 180, because together, they are supplementary. So, I could write that X
+ 90 + Y is going to be equal to 180 degrees. If you want, you can subtract
90 from both sides of that, and you could say look, X
+ Y is going to be equal to 90 degrees. These are algebraically
equivalent statements. So, is equal to 90 degrees. And how can we use this to fill out some of the other angles
in these triangles. Well, let's see X + this angle down here has to be equal to 90 degrees. Or you could say X + 90
+ what is going to be equal to 180. I'm looking at triangle
CBA right over here. The interior angles of a
triangle add up to 180. So, X + 90 + what is equal to 180. Well, X + 90 + Y is equal to 180. We already established that. Similarly over here. Y + 90 + what is going to be equal to 180. Well, same argument. We already know. Y + 90 + X is equal to 180. So, Y + 90 + X is equal to 180. And so, notice we have now established that triangle ABC and triangle EDA, all of their interior angles, their corresponding interior
angles are the same, or that their three
different angle measures all correspond to each other. They both have an angle of X. They both have a measure of X, they both have an angle of measure Y, and they're both right triangles. So, just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates. We know that triangle EDA is similar to triangle ABC. And so, that tells us that
the ratio of corresponding sides are going to be the same. And so, for example, we know. Let's find the ration
of corresponding sides. We know that the ratio
of let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So, the ratio of CB over BA is going to be equal to... Well, the corresponding side to CB, it's the side opposite the X degree angle right over here. So, the corresponding
side to CB is side AD. So, that's going to be equal to AD over, what's the corresponding side to BA? Well, BA is opposite the Y degree angle. So, over here the
corresponding side is DE. AD over DE. Let me do that same color. Over DE. And so, this right over,
we saw from the beginning, this is the slope of L. So, slope of L. And how does this relate
to the slope of M? Notice, the slope of M is
the negative reciprocal of this. You take the reciprocal,
you're going to get DE over AD, and then you have to take
this negative right over here. So, we could write this
as the negative reciprocal of slope of M. Negative reciprocal of M's slope. And there you have it. We've just shown that if we assume L and M are perpendicular, and we setup these similar triangles, and we're able to show, that the slope of L is
the negative reciprocal of the slope of M.