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
fardeen,
Now that you have all of them in the same variable, just calculate and find x minimum using derivatives. For found, calculate using the line equation.
If you're not supposed to use calculus, then inform us, for then a different approach will be needed.
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:Now you square both sides:
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,