$A(4,2)$ and $B(2,4)$ are 2 given points and the point P on the line $3x+2y+10=0$ is given then which of the following is or are true:
(a) $(PA+PB)$ is minimum when $P(-14/5,-4/5)$
(b) $(PA+PB)$ is maximum when $P(-14/5,-4/5)$
(c) $|PA-PB|$ is minimum when $P(-22,28)$
(d) $(PA-PB)$ is maximum when $P(-22,28)$

The only way I can think of solving this question is by first figuring out $y=PA+PB$ and then finding the point P when $dy/dx=0$. Then by substituting any other value of $P$ I identify whether the point I figure out was the maxima or the minima. Then I repeat the whole procedure for $y=PA-PB$.

But this seems to be a horribly wrong procedure and I am searching for a shorter method to solve this question. It would be great if someone could help.

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    $\begingroup$ With no other context - and if my only goal was to answer this one question correctly - I would find $P_1A, P_1B, P_2A, P_2B$ for $P_1 = (-14/5, -4/5)$ and $P_2 = (-22, 28)$ and just use process of elimination. $\endgroup$ – The Chaz 2.0 Apr 11 '17 at 13:08
  • $\begingroup$ @TheChaz2.0 makes sense! Why would you go for such a long process. Thank you! $\endgroup$ – Osheen Sachdev Apr 11 '17 at 13:09
  • $\begingroup$ Just btw this is a more than one correct type... $\endgroup$ – Osheen Sachdev Apr 11 '17 at 13:12
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    $\begingroup$ HINT: $PA+PB$ doesn't have a maximum, $|PA-PB|$ has minimum $0$. $\endgroup$ – Aretino Apr 11 '17 at 13:35
  • $\begingroup$ What you are looking for is the point of contact to an ellipse with focii as the two given points and tangent to the given line. $\endgroup$ – Sawarnik Apr 12 '17 at 8:41

I think this diagram should help you.

enter image description here


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