Question 1116921
A PICTURE IS USUALLY HELPFUL:
The line with equation {{{4x-3y+16=0}}} <--> {{{y=(4/3)x+16/3}}} has slope {{{4/3}}} .
It is easy to see that it passes through point(-4,0).
I can plot points of that line by adding {{{3}}} to the x-coordinate
(moving 3 spaces to the right),
and adding {{{4}}} to the y-coordinate
(moving 4 spaces up) from any point on the line, including point (-4,0).
That allows me to graph that line, along with point {{{C(-9,10)}}} .
{{{drawing(300,300,-11,9,-4,16, grid(1),
red(circle(-9,10,0.3)),blue(circle(-4,0,0.3)),
blue(circle(-1,4,0.3)),blue(circle(2,8,0.3)),
blue(circle(5,12,0.3)),blue(line(-7,-4,8,16)),
locate(-9.8,10,red(C))
)}}}
 
As the line with equation {{{4x-3y+16=0}}} <--> {{{y=(4/3)x+16/3}}} and slope {{{4/3}}} .
as it is the perpendicular bisector of segment CD,
it is perpendicular to the line CD containing segment CD, and points C and D.
Perpendicular lines have slopes whose product is {{{-1}}} ,
so the slope {{{m}}} of the line containing points C and D is such that
{{{m(4/3)=-1}}} --> {{{m=(-1)(3/4)=-3/4}}} .
That slope would allow me to plot points of line CD (and graph line CD),
by adding {{{4}}} to the x-coordinate
(moving 4 spaces to the right),
and adding {{{-3}}} to the y-coordinate
(moving 3 spaces down) from any point on the line, including point C(-9,10).
The points added and line CD are shown in green below.
{{{drawing(300,300,-11,9,-4,16, grid(1),
red(circle(-9,10,0.3)),blue(circle(-4,0,0.3)),
blue(circle(-1,4,0.3)),blue(circle(2,8,0.3)),
blue(circle(5,12,0.3)),blue(line(-7,-4,8,16)),
green(circle(-5,7,0.4)),green(circle(-1,4,0.4)),
green(circle(3,1,0.4)),green(circle(7,-2,0.4)),
green(line(-13,13,11,-5)),locate(-9.8,10,red(C)),
locate(6.2,-2,green(D))
)}}} .
From point to point on each line, I moved
{{{3}}} units in one direction and {{{4}}} perpendicular to that direction,
for a distance (along the line) of {{{sqrt(3^2+4^2)=5}}} ,
so I see that point C is at a distance {{{10}}} from the blue perpendicular bisector line,
so the point at distance {{{10}}} on the other side of that line is point {{{highlight(D(7,-2))}}} . 
 
PROBABLY "FORMULAS" AND EQUATIONS ARE EXPECTED TO BE SHOWN:
The line with equation {{{4x-3y+16=0}}} <--> {{{y=(4/3)x+16/3}}} and slope {{{4/3}}} .
As it is the perpendicular bisector of segment CD,
it is perpendicular to the line CD containing segment CD, and points C and D.
Perpendicular lines have slopes whose product is {{{-1}}} ,
so the slope {{{m}}} of the line containing points C and D is such that
{{{m(4/3)=-1}}} --> {{{m=(-1)(3/4)=-3/4}}} .
 
Knowing the coordinates of {{{C(-9,10)}}} and the slope of the line,
we can write the equation of line CD in point-slope form as
{{{y-10=(-3/4)(x-(-9))}}} or {{{y-10=(-3/4)(x+9)}}} .
Multiplying both sides of the equal sign times {{{4}}} , and rearranging, we get
{{{4y-40=-3x-27}}} and {{{3x+4y-13=0}}} .
 
The intersection of linr CD and the perpendicular bisector of segment CD
is the midpoint of CD, and is given by
{{{system(3x+4y-13=0,4x-3y+16=0)}}} .
Adding up the first equation times 3 plus the second times 4, we get
{{{25x+25=0}}} --> {{{x=-1}}} ,
and substituting that into {{{4x-3y+16=0}}} , we get
{{{-4-3y+16=0}}} --> {{{-3y+12=0}}} --> {{{y=4}}} .
So, the midpoint of CD, {{{M(x[M],y[M])}}} has {{{x[M]=-1}}} and {{{y[M]=4}}} .
The coordinates of the midpoint of a segment CD, with {{{M(x[C],y[C])}}} and {{{M(x[D],y[D])}}} are given by
{{{x[M]=(x[C]+x[D])/2}}} and {{{y[M]=(y[C]+y[D])/2}}} .
Substituting the known coordinates of C and M,
{{{-1=(-9+x[D])/2}}} --> {{{-2=-9+x[D]}}} --> {{{highlight(x[D]=7)}}}
and
{{{4=(10+y[D])/2}}} --> {{{8=10+y[D]}}} --> {{{highlight(y[D]=-2)}}} .