identify a point in the triangle when given other vertices?

two vertices of right triangle ABC are A(-2,6) and C(7,3).

If the right angle is at vertex A and vertex B is on the x-axis,

identify the coordinates of point B

Let \(\vec{A} = \binom{-2}{6}\)
Let \(\vec{B} = \binom{x}{0}\)
Let \(\vec{C} = \binom{7}{3}\)

\(\begin{array}{rcll} \vec{AC} = \ ? \\ \vec{AC} &=& \vec{C} - \vec{A} \\ \vec{AC} &=& \binom{7}{3} - \binom{-2}{6} \\ \vec{AC} &=& \binom{7+2}{3-6} \\ \vec{AC} &=& \binom{9}{-3} \\ \end{array}\)

\(\begin{array}{rcll} \vec{AB} = \ ? \\ \vec{AB} &=& \vec{B} - \vec{A} \\ \vec{AB} &=& \binom{x}{0} - \binom{-2}{6} \\ \vec{AB} &=& \binom{x+2}{-6} \\ \end{array}\)

\(\triangle ABC\) is a right triangle then \(\vec{AC}\cdot \vec{AB} = 0\)

\(\begin{array}{rcll} \mathbf{ \vec{AC}\cdot \vec{AB} } & \mathbf{=} & \mathbf{0} \\ \binom{9}{-3} \cdot \binom{x+2}{-6} &=& 0 \\ 9\cdot(x+2) + (-3)\cdot (-6) &=& 0 \\ 9x+18+18 &=& 0 \\ 9x+ 36 &=& 0 \quad & \quad :9 \\ x+ 4 &=& 0 \\ \mathbf{x} & \mathbf{=} & \mathbf{-4} \\ \end{array} \)

\(\mathbf{\vec{B} = \binom{-4}{0} } \)

Here's another way without using vectors.....let B  = (x, 0)

If the right angle is at A.....the hypotenuse is BC....and this distance  is  just  

sqrt [ (7 - x)^2 + (3 - 0)^2 ] =

sqrt [ 49 - 14x + x^2 + 9 ]   =  sqt [ 58 - 14x + x^2 ]

And AB  forms one of the legs...and its length is just    sqrt [ (-2 - x)^2 + (6 -0)^2 ] =  

sqrt [ ( 4 + 4x + x^2 + 36 ] =

sqrt [ 40 + 4x + x^2 ]

And AC  is the other leg....and its  length is just  sqrt [ (-2 - 7)^2 + (6 -3)^2 ] =

sqrt [ 81 + 9 ]  = sqrt [90]

And by the Pytahgorean Theorem.........AB^2 + AC^2  = BC^2   ....so...

[ 40 + 4x + x^2 ]  +  90    =  [ 58 - 14x + x^2 ]   simplify

130 + 4x  =  58 - 14x          subtract 4x, 58 from both sides

72 = -18x             divide both sides by -18

-4  = x       so ....B  = (-4, 0)

Obviously.....vectors make the process easier.....!!!!