# $A(4,2)$ and $B(2,4)$ are 2 given points and the point P on the line $3x+2y+10=0$. Find $P$ for which $PA+PB$ is maximum

$A(4,2)$ 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.

• 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. – The Chaz 2.0 Apr 11 '17 at 13:08
• @TheChaz2.0 makes sense! Why would you go for such a long process. Thank you! – Osheen Sachdev Apr 11 '17 at 13:09
• Just btw this is a more than one correct type... – Osheen Sachdev Apr 11 '17 at 13:12
• HINT: $PA+PB$ doesn't have a maximum, $|PA-PB|$ has minimum $0$. – Aretino Apr 11 '17 at 13:35
• 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. – Sawarnik Apr 12 '17 at 8:41