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.

  • $\begingroup$ Looks straight forward, if tedious, to me. The equation of line MN is y= (3/5)x+ 19/5. Writing P as (a, b), distance MP is $\sqrt{a+ 3)^2+ (b- 2)^2}$ and PN is $\sqrt{(a- 2)^2+ (b- 5)^2}$. So the problem is to maximize $\left|\sqrt{(a+ 3)^2+ (b- 2)^2}- \sqrt{(a- 2)^2+ (b- 5)^2}\right|$ with the condition that $b= (3/5)a+ 19/5$. You could either replace b by that in the quantity to be maximized or use "Lagrange multipliers". $\endgroup$ – user247327 May 3 '17 at 12:03

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, $$MN' \ge MP - N'P.$$


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