1. ## Analytic geometry

Coordinates of point A is (1,3) and B is (5,-3).There is a point P
moving over line y=x+4. Find minimum value of |AP|+|PB|.
I used the distance of a point to a line formula but I found the

2. ## Re: Analytic geometry

$|AP| = \sqrt{(x-1)^2+(x+4-3)^2}$
$|PB| = \sqrt{(5-x)^2+(-3-x-4)^2}$

The sum is:
$\sqrt{2x^2+2}+\sqrt{2x^2+4x+74}$

This is minimized when $x=-\dfrac{1}{7}$ and the result is 10, not 6.

3. ## Re: Analytic geometry

Btw, suppose $\sqrt{2x^2+4x+74}+\sqrt{2x^2+2}=k$

Multiply both sides by the conjugate:
$\sqrt{2x^2+4x+74}-\sqrt{2x^2+2}=\dfrac{4x+72}{k}$

Adding these together and squaring both sides gives:

$4(2x^2+4x+74)=\left(k+\dfrac{4x+72}{k}\right)^2$

4. ## Re: Analytic geometry

To minimize $PA+PB$, $P$ must be at the intersection of the line $y=x+4$ with the line $BC$
where $C$ is the symmetric of $A$ with respect to the line $y=x+4$

$C(-1,5)$

$B(5,-3)$

minimum distance = $\sqrt{6^2+8^2}=10$

Many Thanks.

6. ## Re: Analytic geometry

Then we could use the symmetry of a point to a line and find C. Multiplication of the slopes will be -1. Then find CB by using distance formula.

7. ## Re: Analytic geometry

P will be(0,4) from the midpoint formula we get C.