Is this a correct way to solve this high school coordinate geometry question? Here's the question:

Given point $A$: $(-3;-1)$
  Given point $B$: $(3;7)$
Given point $Z$: $(x;0)$
Find the $x$ coordinate of point $Z$ so that the angle of view of AB segment is $90$ degrees at point Z, so Point Z has a $90$ degrees interior angle.
The answer is $4$ and $-4$.

Here's the way I did it, is it correct?


*

*$AB$ segment has length of $10$.

*So to make $ABZ$ a right triangle, one method we can use is that $AZ$ segment and $BZ$ segment must have the length of $\sqrt{50}$, because then $\sqrt{50}^2 + \sqrt{50}^2 = 10^2$. (Which is the length of $AB$)

*So in that case triangle $ABZ$ would be an isosceles right angled triangle and indeed there would be a $90$ degrees interior angle at point $Z$.

*Then, I used distance formula to calculate the $x$ coordinate of point $Z$ so that $AZ$ and $BZ$ segments would have a length of $\sqrt{50}$.


So that's how I got $4$ and $-4$, and these are the correct answers.
I know there are many possible solutions to solve this problem, but is my method correct?
 A: First of all, the 
angle of view of a line segment are two arcs and Z are on these arc and also it is on a line, that is the $y=0$ line. Your task is to figure out the formula for the arcs and then find the intersections of $y=0$ on it. Edit: hint, you need to use the Central Angle Theorem to find the centre of the arc/circle.
I do not understand why $ABZ$ is a right triangle in the first place. 
You claim  

So to make $ABZ$ a right triangle, one method we can use is that $AZ$ segment and $BZ$ segment must have the length of $\sqrt{50}$, because then $\sqrt{50}^2 + \sqrt{50}^2 = 10^2$. (Which is the length of $AB$)

This is wrong. If AZ and BZ equal length and the angle at Z is 45 degrees then the other two angles of triangle is (180-45)/2.
A: your method is correct. Good job sorry it took so long,  it's hard to review things not in math Jax I highly recommend you learn
A: It is incorrect
First of all, 
$Z(x, 0)$
Now for every possible value of x, a line can be made that is intersecting line AB at $45 $ degree.
i think you mean point $A$ and $B$ is seen at $45$ degrees.
i got the answer this way.
$m_1$ = $\frac{7+1}{3+3}$
$m_1$ = $\frac{4}{3}$
$tan 45$ = $\frac{-m_1 + m_2}{1 + m_1*m_2}$
$1 + m_1*m_2$ = ${ + m_2 -m_1}$
$1 + \frac{4}{3}*m_2$ = ${+m_2 -\frac{4}{3}}$
$3 + 4*m_2$ = ${-4 + 3*m_2}$
$m_2$ = ${-7}$
$m_2$ = $-7$
$\frac{7-y}{3-x}$ = $-7$
$7x +y -28$ = $0$
Now it is given that y = $0$
$7x - 28 $ = $0$
$x$ = $+4$
Do the same for $A$.
A: So all the fuzz here resulted from some unsharp definitons (where is the angle, what angle).

You can try out different $x$ values here.
