The task is:
Given points $M(-3,2)$ and $N(2,5)$. Find a point $P$ on $OY$ that $|MP-NP|$* is maximum.(You are not allowed to use anything(vectors) except the general equation of straight line and the definition of the distance between two points)
I found a symmetric point to $N$ : $N'(-2,5)$. And as I understood the next step is to find the line $l$ between $N'$ and $M$ : $3x-y+11=0$ and than intersection $l \cap OY=P(0,11)$.But why $|MP-NP|$ is maximum I still can't understand.
(*MP-the distance between M and P)
Thanks in advance.
First, note that the distances $NP$ and $N'P$ are equal (check why).
Now, you can draw a triangle with sides $N'P$, $MP$ and $MN'$. You always have $$MN' + N'P \ge MP,$$ where the equality is true only when the three sides lie on the same line (see triangle inequality). That's only possible in the case you have described. And, thus, $$MN' \ge MP - N'P.$$