Question 983311
(After drawing a graph on a chalkboard and thinking very carefully on this, a solution method began to come to me...)



Draw and label a graph of the situation as completely as you can do.
Show your points G(-4,-3) and H(4,9); and judge what two points (x,y) on the x-axis would form two possible points P(x,y), so that you can draw lines.


The four points  (two different "P"), and G and H will form a trapezoidal figure.  Place your two "P" points so that PG and PH LOOK AS THOUGH they meet at right angle, or are perpendicular.


You will use the slopes of GP and PH.
Perpendicular lines will have slopes which are negative reciprocals of each other.
(You really need to know formula for slope of a line for this!)


Setup this table.


SLOPE________________EXPRESSION
Line GP______________{{{(y+3)/(x+4)}}}
Line PH______________{{{(y-9)/(x-4)}}}


Again, these slopes must form a product of {{{-1}}}.
Next setup the equation for this:
{{{highlight_green(((y+3)/(x+4))((y-9)/(x-4))=-1)}}}

I here omit a few steps...
but you should get a step continuing with
{{{y^2-6y-27=-x^2+16}}}
and then continuing,
{{{y^2-6y+x^2=27+16}}}
{{{x^2+y^2-6y=43}}}
You also MUST know about Completing the Square  (maybe, maybe not...)
{{{x^2+y^2-6y+9=43+9}}}
{{{highlight_green(x^2+(y-3)^2=52)}}}


That looks like it will be complicated to find your point P(x,y); but remember you are looking for possible points ON THE x-AXIS, and therefore {{{highlight(highlight(y=0))}}}.


NOW, what is x?