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I was solving a problem that required me to find some points in the plane perpendicular to a segment that I already had. Since I am applying the fact that the dot product of the vector I already have and the new vector I am trying to discover is zero (assuming that they are perpendicular), and considering that I also know the length of this new vector I am trying to discover, I wrote a system of equations:

$$ 550 x + 350 y = 0 \\ 50 = \sqrt{ (x - 150)^2 + (y - 50)^2 } $$

Before I started solving by myself, I checked the solution in Wolfram Alpha, and yes. The results are exactly the points I need to create my new segment, perpendicular to the one I had.

But when solving "by hand", I could do the first equation, but when trying to solve the second one, I arrived to a moment when I get as a result a square root of a negative number, and I have no idea about how to continue from there.

I am solving the lower equation using:

$$ (a-b)^2 = a^2-2ab+b^2 $$

Is this the right method or I have to do something else?

I am not very good at this, so I am trying to read as much as I can to understand how this works, but for the moment I am having trouble. Any comment regarding this problem will be much appreciated.

Thanks!

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1 Answer 1

up vote 7 down vote accepted

$$(550x)+ (350y) =0 \quad \iff \quad 350 y = -550 x \quad\iff \quad y = - \frac{550}{350} = \left(-\frac {11}{7}x\right)\tag{1}$$

Now substitute $\quad y = \left(-\dfrac{11}{7}x\right)\quad$ into your equation:

$$50 = \sqrt { { (x - 150) }^{ 2 } + { (y- 50) }^{ 2 } }\tag{2} $$

and then solve for $x$.


I suspect that in your computations, you must have made an error, if you obtained a negative value under the square-root:

Since both terms in the expression $$\sqrt { { (x - 150) }^{ 2 } + { \left[\left(-\frac{11}{7}x\right) - 50\right] }^{ 2 } }$$ are first squared (hence positive), and then summed, you should have the square root of a positive number.

Note: if you square both sides of your equation $(2)$ to solve for $x$, you will get two solutions for $x$, but you want only the solution(s) which is defined when inserting into equation $(2)$. By squaring each side of the equation to solve for $x$, you may introduce a "solution" to the quadratic that in fact may not solve the original equation $(2)$.

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Thanks, that's what I was doing, but in the last step of solving for x, I ran into the square root of a negative number. That's what stopped me from continuing... –  telex-wap Dec 10 '12 at 20:06
    
telex-wap: see my expanded answer... –  amWhy Dec 10 '12 at 20:20
    
@telex-wap: A scan of the equation (essentially a quadratic) shows that both solutions, one of which is probably unwanted, are positive. So I expect there was a minus sign error, all too easy to make, at least for me. –  André Nicolas Dec 10 '12 at 20:24

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