Point for minimum distance

Question

Find the coordinates of the point P on the line $d : 2x − y − 5 = 0$ for which the sum $AP + PB$ is minimum, when $A(−7, 1)$ and $B(−5, 5)$. Can somebody help me, please?

Answer 1

Well, a set of points on the line $\text{d}$ is given by:

$$\text{P}\left(x,\text{y}\right)=\text{P}\left(x,2x-5\right)\tag1$$

Using the distance formula, we can find:

• $$\left|\text{AP}\right|=\sqrt{\left(x-\left(-7\right)\right)^2+\left(2x-5-1\right)^2}\tag2$$
• $$\left|\text{BP}\right|=\sqrt{\left(x-\left(-5\right)\right)^2+\left(2x-5-5\right)^2}\tag3$$

Now, find:

$$\frac{\text{d}}{\text{d}x}\left(\left|\text{AP}\right|+\left|\text{BP}\right|\right)=0\tag4$$

Which will give $x=2$, and $\text{y}=2\cdot2-5=-1$ so $\text{P}\left(2,-1\right)$.

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Using my comment, you can see:

$$\frac{\text{d}}{\text{d}x}\left(\sqrt{\left(x-\left(-7\right)\right)^2+\left(2x-5-1\right)^2}\right)=\frac{\left(x-\left(-7\right)\right)+2\left(2x-5-1\right)}{\sqrt{\left(x-\left(-7\right)\right)^2+\left(2x-5-1\right)^2}}\tag5$$

And:

$$\frac{\text{d}}{\text{d}x}\left(\sqrt{\left(x-\left(-5\right)\right)^2+\left(2x-5-5\right)^2}\right)=\frac{\left(x-\left(-5\right)\right)+2\left(2x-5-5\right)}{\sqrt{\left(x-\left(-5\right)\right)^2+\left(2x-5-5\right)^2}}\tag5$$

Answer 2

Segment $AB$ is parallel to the line. Therefore $AP+PB$ is minimum if $\vec{P}=\frac{\vec{A}+\vec{B}}{2}+t\{2\hat{i}-\hat{j}\}$.

Substitute this to the equation line to obtain $t=4$. Thus $\vec{P}=\{2\hat{i}-\hat{j}\}$

Answer 3

Any point $P$ on that line would have a general form of $(x, 2x-5)$

Then you would evaluate the following:

$|PA|^2=(x+7)^2+(2x-6)^2$

$|PB|^2=(x+5)^2+(2x-10)^2$

Now the question asks us to minimize $PA+PB$, which would involve the square roots of these signs. But here's a better idea.

$\operatorname{min}(|PA|+|PB|) \equiv \operatorname{min}\left ((|PA|+|PB|)^2 \right)$

$\equiv \operatorname{min}(PA^2+PB^2+2 \cdot |PA| \cdot |PB|)$

$\equiv \operatorname{min}(PA^2+PB^2)$

So we just minimize the sum of their squares.

Which in this case totals to:

$10x^2-40x+210=10(x^2-4x+21)$

This is a parabola so finding it's minimum should give us the minimum for that abscissa. Doing that:

$2x-4=0 \implies x=2$

So the point $P$ that gives minimum expression is $(2, -1)$. Peace.

Fun fact: $PAB$ is isosceles, which if I knew geometrically was true I would've gone a different route, of making an isosceles triangle instead.