System of equations is valid, but cannot solve, since I find a square root of negative number 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!
 A: $$(550x)+ (350y) =0 \quad \iff \quad 350 y = -550 x \quad\iff \quad y = - \frac{550}{350} = \left(-\frac {11}{7}x\right)\tag{1}$$
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 $(2)$.
