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There is the second problem of the day that I have been stuck on for quite some time, and I am having trouble examining how to evaluate this equation to simple form.

Prove the lines given by the functions

$$\begin{align*} f(x) & {} = dx + b \\ g(x) & {} = ex + c \\ \end{align*}$$

are perpendicular, if $de = -1$ verify that the squares of lengths of sides of the triangle that have vertices of $(0,0), (1,d) \textrm{ and } (1,e)$. Also, can you guys explain why we can say that the lines will intersect at (0,0)?

Thanks a lot! I will have to study this explanation so please be discrete!

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Well, if $de=-1$ it means that $d$ is a negative reciprocal of $e$ or visa versa, ex.g. $d=4$ and $e=-1/4$. Also, $d$ and $e$ are the slopes of the lines. So, when the product of the slopes of lines is $-1$ (the slopes are negative reciprocals) it means they are perpendicular. Simple as that. The second part of the questions about the triangle I dont quite understand. You need to edit the post. –  Koba Sep 12 '13 at 3:41
Well the question that I am given is this: Prove the lines given by the functions: f(x)=mx+b g(x)=nx+c are perpendicular if mn = -1 by computing the squares of the lengths of the sides of the triangle with vertices (0,0), (1,m), and (1,n). Make sure you explain why we can assume the lines intersect at (0,0). That is word for word. I don't exactly know how to change it up anymore :/ –  user2581649 Sep 12 '13 at 3:47

2 Answers 2

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If you are simply asked to prove that the lines are perpendicular by computing the squares of the sides of a triangle with vertices $(0,0), (1,m)$, and $(1,n)$ you just have to compute the squares of the sides of the triangle with vertices $(0,0), (1,m)$, and $(1,n)$.enter image description here

I added an ugly picture for clarity. There you go, you have a triangle ABC with aforementioned vertices. So, now the side AC=$\sqrt{1^2 + m^2}$ by the Pythagorean Theorem. Similarly, BC=$\sqrt{1^2 + n^2}$. Now, you need to add the squares of the sides of ABC, i.e. $AC^2 + BC^2$ which is equal to $(AB)^2$ which is equal to $(m+n)^2$. Solve the equality $(AC)^2 + (BC)^2=(m+n)^2$ and you will get that $mn=-1$.

Why can you assume that the lines intersect at (0,0)? Because there are no restrictions on $b$ and $c$, so we can assume they are zero. And lines of the form $f(x)=ax$ pass through the origin.

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There is no reason at all why they should intersect at (0,0). Take for example f(x) = x + 1 and g(x) = -x + 1. They are perpendicular but intersect at (0,1). However, you were not asked if they intersect at (0,0).

You are asked to find something about the sides of the triangle which in this case would be (0,0), (1,1) and (1, -1). What is it you are being asked to show? If you were clearer about that you might find the problem easier.

I suspect you are asked to show that the Pythagorean theorem holds for this triangle (it better, because this is a right triangle).

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I am simply asking to prove the lines given by the functions that I specified are perpendicular if the conditions specified afterwards are met. –  user2581649 Sep 12 '13 at 3:33

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