# Minimum of PA + PB

#### jacks

##### Well-known member
Find a point $P$ on the line $3x+2y+10=0$ such that $PA+PB$ is minimum given that $A$ is

$(4,2)$ and $B$ is $(2,4)$

My Try:
Let Coordinate of point $P$ be $(x,y)$. Then $PA = \sqrt{(x-4)^2+(y-2)^2}$ and $PB = \sqrt{(x-2)^2+(y-4)^2}$

Now Let $f(x,y) = \sqrt{(x-4)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-4)^2}$

Now How can i Minimize $f(x,y)$

Help me

Thanks

#### MarkFL

Staff member
Re: Minimum of PA+PB

Is this a question given in a calculus course?

#### MarkFL

Staff member
Re: Minimum of PA+PB

I have taken the liberty of moving this thread to our Calculus subforum. A review of your recent posts seems to suggest that you were in Calculus II last term and so I feel this is most likely an optimization problem with a constraint meant to be an application of Lagrange multipliers.

So, I would identify (as you did) the objective function as:

$$\displaystyle f(x,y)=\sqrt{(x-4)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-4)^2}$$

Subject to the constraint:

$$\displaystyle g(x,y)=3x+2y+10=0$$

Can you proceed from here?

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Re: Minimum of PA+PB

Find a point $P$ on the line $3x+2y+10=0$ such that $PA+PB$ is minimum given that $A$ is $(4,2)$ and $B$ is $(2,4)$My Try: Let Coordinate of point $P$ be $(x,y)$. Then $PA = \sqrt{(x-4)^2+(y-2)^2}$ and $PB = \sqrt{(x-2)^2+(y-4)^2}$Now Let $f(x,y) = \sqrt{(x-4)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-4)^2}$Now How can i Minimize $f(x,y)$Help meThanks
This is best solved using Euclidean geometry.
Let $l$ be a given line and $A$ and $B$ be two points given on the plane.We need to find a point $P$ on the line $l$ such that $|PA|+|PB|$ is minimum.

If $A$ and $B$ are on different sides of the line $l$, then simply join $A$ and $B$ by a straight line $m$ and the point of intersection of $l$ with $m$ is the required point $P$.

If $A$ and $B$ are on different sides of the line $l$ then reflect $B$ about the line $l$.Say $B'$ is the reflection of $B$ about $l$.Join $A$ and $B'$ through a straight line $m'$. The point of intersection of $m'$ with $l$ is the required point $P$.

If you need a justification why the above works you can ask me but I suggest you try justifying it yourself first. It is not hard. You simply need to make use of the fact that the shortest path between two points on the Euclidean plane is through the straight line passing through the points.

#### MarkFL

Staff member
Re: Minimum of PA+PB

Nice one, caffeinemachine!

I can't believe I forgot the "trick of reflection" for this type of problem.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Re: Minimum of PA+PB

Nice one, caffeinemachine!

I can't believe I forgot the "trick of reflection" for this type of problem.
Thanks

#### johng

##### Well-known member
MHB Math Helper
Re: Minimum of PA+PB

Here's a diagram to go with caffeinemachine's excellent solution: