Vikas TU
Last Activity: 8 Years ago
Writing the P coordinates in parametric form that is in terms of x and y respectively, we get,
P (x , (4x – 9)/3)
Distance PA => root(x^2 + ((4x – 12)/3)^2)
Distance PB = > root( (x-2)^2 + ((4x-9)/3)^2)
Now, let M = |PA – PB|
= |root(x^2 + ((4x – 12)/3)^2) – root( (x-2)^2 + ((4x-9)/3)^2)|
To maximize this, root( (x-2)^2 + ((4x-9)/3)^2) should be minimized.
and for that root( (x-2)^2 + ((4x-9)/3)^2) = 0
that is (x-2)^2 = 0 and also (4x-9)/3)^2 = 0
x = 2 and x = 9/4
Hence, y = -17/3 and y = -6 respectively.
Tese are the final coordinates.
