[SOLVED] |AP – BP| is minimum for...?

**fardeen_gen** · Oct 27th 2008, 06:52 AM

Let A(-2, -1) & B(3, 2) be two fixed points and P be a point on the line 3x + y = 1, for which location of P, |AP – BP| is minimum?

A) (2,-5)
B) (0,1)
C) (1, -2)
D) none

**Rafael Almeida** · Oct 27th 2008, 07:28 AM

fardeen,

$P = (x,3x+1)$
$AP = d(A,P) = \sqrt{(x + 2)^2 + (3x + 2)^2}$
$BP = d(B,P) = \sqrt{(x - 3)^2 + (3x - 1)^2}$

Now that you have all of them in the same variable, just calculate $|AP - BP|$ and find x minimum using derivatives. For $x_{min}$ found, calculate $y_{min}$ using the line equation.

If you're not supposed to use calculus, then inform us, for then a different approach will be needed.

**fardeen_gen** · Oct 27th 2008, 07:35 AM

No calculus. Only coordinate geometry and normal geometry.

**fardeen_gen** · Oct 27th 2008, 07:57 AM

Also I couldn't understand how do we get P(x , 3x + 1). Can you please explain?

**Rafael Almeida** · Oct 27th 2008, 08:14 AM

fardeen,

1) A point P is denoted by an ordered pair (x,y). Since your line equation is 3x + y = 1, then y = 1 - 3x (I obviously got the signs wrong in my last post, I'm sorry). Since you now know y in terms of x, then P = (x,1-3x).

2) Now that I though a little further into the question, consider this approach: first notice that (AP - AB) is able to assume every value in the real line. So the minimum value of the modulus must be zero. This obviously happens when P is equidistant of both A and B.

So your condition is:

$d(P,A) = d(P,B)$
$\sqrt{(x + 2)^2 + (3x + 2)^2} = \sqrt{(x - 3)^2 + (3x - 1)^2}$

Now you square both sides:

$(x + 2)^2 + (3x + 2)^2 = (x - 3)^2 + (3x - 1)^2$

This is a quadratic equation. Solve it for x and you will find the value of x for which distances are equal, and because of such the difference is zero.

Hope this helps,

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