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:?

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)#

#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)#