Question 1063151
We can apply the "distance formula"
to set up and solve a quadratic equation
that would allow us to find the y-coordinate(s)
of the point(s) (9,y), at distance 10 from (3,-2), if any.
That may be what was intended.
We can also understand and reason more,
while doing less calculations.
All the points with an x-coordinate of 9
are on the vertical line x=9.
All the points at a distance of 10 from point (3,-2)
are on the circle of radius 10 centered at (3,-2).
Since the distance between point (3,-2], with x=3,
and line x=9 is 9-3=6,
there will be two points meeting both requirements,
at the intersection of line and circle.
Going horizontally from (3,-2) to (9,-2),
continuing vertically for a distance b to one of those points,
and returning on a straight line to (3,-2),
we "draw" a right triangle.
The vertical distance, b, according to the Pythagorean theorem,
satisfies {{{b^2+6^2=10^3}}} .
We can solve that to find x.
We can also realize that the very popular 3-4-5 right triangle,
with sides measuring 3, 4, and 5,
is similar to our trisngle wicj has sides measuring
{{{2*3=6}}} , {{{2*5=10}}} , and
{{{2*4=b}}} , so {{{b=8}}} ,
and the points have
{{{y=-2-8=highlight(-10)}}} and {{{y=-2+8=highlight(6)}}} .