Question 1032691
(a) is easy.
A picture is not needed, but I will add a picture so you can visualize it:
{{{drawing(300,600,-2,8,-2,14,
grid(0),circle(-1,3,0.1),
circle(3,9,0.1),circle(5,12,0.1),
green(triangle(-1,3,3,3,3,9)),
red(triangle(3,9,5,9,5,12)),
locate(-1.5,3.2,A),locate(2.5,9.3,B),locate(4.9,12.5,C),
locate(0.3,3.3,green(x[B]-x[A])),locate(3.1,6.2,green(y[B]-y[A]))
)}}}

Given points {{{A(x[A],y[A])}}} and {{{B(x[B],y[B])}}} , the slope of AB is
{{{(y[B]-y[A])/(x[B]-x[A])}}}
For {{{A(-1,3)}}} and {{{B(3,9)}}} , the slope of AB is
{{{(9-3)/(3-(-1))=6/(3+1)=6/4=3/2}}} .
For {{{B(3,9)}}} and {{{C(5,12)}}} , the slope of BC is
{{{(12-9)/(5-3)=3/2}}} .
Since AB and BC have the same slope, they are either parallel or the same line.
As they have point B in common, AB and BC are the same line.
 
(b) ONE WAY TO GO ABOUT IT (ugly, but probably the expected way):
If point P is {{{P(x,y)}}}
{{{AP^2=(x-(-1))^2+(y-3)^2=(x+1)^2+y^2-6y+9=x^2+2x+1+y^2-6y+9=x^2+y^2+2x-6y+10}}} ,
{{{BP^2=(x-3)^2+(y-9)^2=x^2-6x+9+y^2-18y+81=x^2+y^2-6x-18y+90}}} , and
{{{CP^2=(x-5)^2+(y-12)^2=x^2-10x+25+y^2-24y+144=x^2+y^2-10x-24y+169}}} .
So {{{AP^2+CP^2=x^2+y^2+2x-6y+10+x^2+y^2-10x-24y+169=2x^2+2y^2-8x-30y+179}}} , and
{{{2bP^2=2(x^2+y^2-6x-18y+90)=2x^2+2y^2-12x-36y+180}}}
So, {{{AP^2+CP^2=2BP^2}}} means
{{{2x^2+2y^2-8x-30y+179=2x^2+2y^2-12x-36y+180}}}
{{{-8x-30y+179=-12x-36y+180}}}
{{{12x-8x+36y-30y=180-179}}}
{{{4x+6y=1}}} 
That is the equation of a straight line, which is the locus of P.
Transforming the equation into slope-intercept form, we get
{{{4x+6y=1}}} --> {{{6y=-4x+1}}} --> {{{y=(-4x+1)/6}}} --> {{{y=(-2/3x)+1/6}}} .
We could also just find the slope, using a formula.
Either way, the slope of the line is {{{-2/3}}} .
If the product of that slope and the slope of AB (found in part (a) is {{{-1}}} ,
then the lines are perpendicular, and it so happens that
{{{(-2/3)(3/2)=-1}}} , so the locus of P is a line perpendicular to AB.
 
ANOTHER WAY (possible, depending on what you have already covered in math classes):
When calculating the slopes of AB and BC,
you may have noticed that for points A, B and C
{{{x[B]-x[A]=2(x[C]-x[B])}}} and {{{y[B]-y[A]=2(y[C]-y[B])}}} .
That tells you that for the distances {{{AB=2BC}}} .
We could calculate those distances, but I only care about their ratios,
so for easier writing, I will rename the distances as {{{BC=c}}} and {{{AB=2c}}} .
You may think of point P as not being on line AB.
However, as it is a moving point, at some point it could be on line AB,
but in that very special case, I would call it point D, and I will say it is a distance {{{d}}} to the other side of A.
I will find {{{d}}} .
The situation would be like this, with the distances:
{{{drawing(800,100,-1.5,6.5,-0.5,0.5,
arrow(-2,0,9,0),circle(-1,0,0.05),locate(-1.1,0.2,D),
circle(0,0,0.05),locate(-0.05,0.2,A),
circle(4,0,0.05),locate(3.95,0.2,B),
circle(6,0,0.05),locate(6.05,0.2,C),
locate(-0.55,0,d),locate(1.9,0,2c),locate(4.95,0,c)
)}}} . (If D is not on the side of A I assumed it to be, {{{d}}} will be a negative value).
Since now P is at D, the equation with the squares is
{{{d^2+(3c+d)^2=2(2c+d)^2}}}
{{{d^2+9c^2+6cd+d^2=2(4c^2+4cd+d^2)}}}
{{{2d^2+9c^2+6cd=8c^2+8cd+2d^2}}}
{{{9c^2+6cd=8c^2+8cd}}
{{{9c^2-8c^2=8cd-6cd}}}
{{{c^2=2cd}}} and since we know that {{{c<>0}}} (even though we did not calculate it), we divide both sides by {{{c}}} and get
{{{c=2d}}} .
Now, what about a point P that is not D?
{{{drawing(800,500,-1.5,6.5,-0.5,4.5,
arrow(-2,0,9,0),circle(-1,0,0.05),locate(-1.1,0.2,D),
circle(-0.8,4,0.05),locate(-0.85,4.2,P),triangle(-1,0,-0.8,4,0,0),
circle(0,0,0.05),locate(0,0.2,A),triangle(4,0,-0.8,4,6,0),
circle(4,0,0.05),locate(3.95,0.2,B),
circle(6,0,0.05),locate(6.05,0.2,C),
locate(-0.55,0,d),locate(1.9,0,4d),locate(4.95,0,2d)
)}}} .
With the Pythagorean theorem (applied to all 3 triangles including side DP), we can easily prove that if P is such that AB and DP are perpendicular, then {{{AP^2+CP^2=2BP^2}}} .
The other way around (proving that if {{{AP^2+CP^2=2BP^2}}} , then AP and DP are perpendicular) I believe requires using the law of cosines, which may be beyond what you have studied.