Coordinate Geometry

TracyStacy:

if A=(2,10) and C=(8,4). Find point B if it lies on the X-axis and AB+BC is a minimum.

I am sorry but I don't know. I will make a point of letting the others know that your question is not answered. sorry.

TracyStacy:

if A=(2,10) and C=(8,4). Find point B if it lies on the X-axis and AB+BC is a minimum.

Let B=(x,0). Use Pythagoras to find the the distance between A and B as a function of x, and to find the distance between C and B as a function of x. Add the two distances - the sum will be a function of x. Differentiate the result with respect to x and set the result to zero. Solve for x.

Alan:

TracyStacy:

if A=(2,10) and C=(8,4). Find point B if it lies on the X-axis and AB+BC is a minimum.

Let B=(x,0). Use Pythagoras to find the the distance between A and B as a function of x, and to find the distance between C and B as a function of x. Add the two distances - the sum will be a function of x. Differentiate the result with respect to x and set the result to zero. Solve for x.

Yes this is a good idea but it is not so easy to do. Have you tried it?

Melody:

Alan:

TracyStacy:

if A=(2,10) and C=(8,4). Find point B if it lies on the X-axis and AB+BC is a minimum.

Let B=(x,0). Use Pythagoras to find the the distance between A and B as a function of x, and to find the distance between C and B as a function of x. Add the two distances - the sum will be a function of x. Differentiate the result with respect to x and set the result to zero. Solve for x.

Yes this is a good idea but it is not so easy to do. Have you tried it?

Yes. A little tedious I'll agree. If I knew how to post a png image here I'd show the solution!

Ok. Best I can think of is to upload the solution as an attachment (PointB.png).

PointB.PNG

Here's a simpler solution.

PointBb.PNG

Hi Alan,
welcome to the Web2.0 forum, we hope you like it here.

I really liked your solutions, especially the second one.
I hope you don't mind, I just want to discuss it a little.
In my diagram you can see that BC = BC' because C' is the reflection of C in the x axis.
The aim is to minimize the distance ABC given that B can be anywhere on the x axis.
Now distance ABC is exactly the same as distance ABC'
140225 shortest distance between 2 points.JPG
So, how can we minimize the distance ABC' ?
Well it all comes down to one fact.
THE SHORTEST DISTANCE BETWEEN 2 POINTS IS A STRAIGHT LINE.
So if B lays on the line AC' then the distance will be least.
A(2,10) and C' (8,-4)
the equation of a line joining two points is
Find the equation of the line joining these 2 points. It is
3y+7x =44
substitute in y=0 because that is the equation of the x axis.
7x = 44
x = 44/7 = 6 and 2/7
so B is the point (6 and 2/7 , 0)

I think that this is really cool.
Thanks Alan for that great answer.

Yes, your approach is essentially the one that provides a proof for my "optical" approach. Very nice!

bump