Question

The coordinates of a point P on the line 2x - y + 5 = 0 such that `|PA - PB|` is maximum, where A is (4,-2) and B is (2, -4) will be:?

A) (11, 27)
B) (-11, -17) (Answer)
C) (-11, 17)
D) (0, 5)

Answer 1

The maximum value of `|PA-PB|` is `AB`

From then on, just solve for the coordinates of P

Explanation:

Consider the triangle `PAB`

From the Triangle Inequality,

`PB+AB>PA`

`AB > |PA - PB|`

Hence the maximum value of `|PA-PB|` is `AB`

Now, just find the coordinates of P given `AB = |PA-PB|`

Let `Q_x` and `Q_y` denote the x- and y-coordinates of an arbitrary point Q

`sqrt((A_x-B_x)^2+(A_y-B_y)^2)=|sqrt((P_x-A_x)^2+(P_y-A_y)^2)-sqrt((P_x-B_x)^2+(P_y-B_y)^2)|`

`2sqrt(2)=abs(sqrt((P_x-4)^2+(P_y+2)^2)-sqrt((P_x-2)^2+(P_y+4)^2))`

Now just solve the above equation and you should get `P_y=P_x-6`

Since P lies on the line y=2x+5, we also know that `P_y=2P_x+5`

After solving the two simultaneous linear equations, you should get `P_x=-11` and `P_y=-17`

Hence P is at `(-11, -17)`

Answer 2

See below.

Explanation:

`p_a=(4,-2)`
`p_b=(2,-4)`

`L-> 2x-y+5=0` or
`L->p_0+mu vec v` with

`p_0 = (0,5)`
`vec v = (1/sqrt(5),2/sqrt(5))`
`mu in RR`

now defining `delta = abs(norm(p-p_a)-norm(p-p_b)` with `p=(x,y)`

we have that `max delta = norm(p_a-p_b)` and this is attained at point `p_s` such that

`p_s = L nn L_(ab)` where

`L_(ab)->p_a+ lambda(p_b-p_a)` with `lambda in RR`

but `p_s` is the solution for

`p_0+mu vec v = p_a+lambda(p_b-p_a)` or

`{(mu/sqrt5=4+lambda(2-4)),(5+mu 2/sqrt5 = -2+lambda(-4+2)):}`

and solving for `mu, lambda` we get

`mu=-11 sqrt5, lambda=15/2` and substituting

`p_s = (-11,-17)`