SOLUTION: Please help me solve this; Given are two points A(-1,3) and B(3,9).(a) Show that C(5,12)is a point of AB. (b) A Point P(x,y) moves in such a way that AP^2+CP^2=2BP^2. Find the equ

Question 1032691: Please help me solve this;
Given are two points A(-1,3) and B(3,9).(a) Show that C(5,12)is a point of AB. (b) A Point P(x,y) moves in such a way that AP^2+CP^2=2BP^2. Find the equation of the locus of P. (c)Show that this locus is a straight line perpendicular to AB.

Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
(a) is easy.
A picture is not needed, but I will add a picture so you can visualize it:

Given points and , the slope of AB is

For and , the slope of AB is
.
For and , the slope of BC is
.
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
,
, and
.
So , and

So, means

That is the equation of a straight line, which is the locus of P.
Transforming the equation into slope-intercept form, we get
--> --> --> .
We could also just find the slope, using a formula.
Either way, the slope of the line is .
If the product of that slope and the slope of AB (found in part (a) is ,
then the lines are perpendicular, and it so happens that
, 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
and .
That tells you that for the distances .
We could calculate those distances, but I only care about their ratios,
so for easier writing, I will rename the distances as and .
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 to the other side of A.
I will find .
The situation would be like this, with the distances:
. (If D is not on the side of A I assumed it to be, will be a negative value).
Since now P is at D, the equation with the squares is

and since we know that (even though we did not calculate it), we divide both sides by and get
.
Now, what about a point P that is not D?
.
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 .
The other way around (proving that if , then AP and DP are perpendicular) I believe requires using the law of cosines, which may be beyond what you have studied.

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