Find the coordinates of the point P on the line d : 2x − y − 5 = 0, for which the sum AP + PB attains its minimum, when A(−7; 1) and B(−5; 5).

Question

I tried to use the Cosine theorem and get that |(PA)^2 + (PB)^2 - (AB)^2|<=|2PAPB|. Can someone explain to me what should I do? Thank you!

Answer 1

hint

The coordinates of the point $ P $, will have the form $ (x,2x-5)$

So

$$AP=\sqrt{(x+7)^2+(2x-5-1)^2}$$ $$=\sqrt{5(x^2-2x+17)}$$ and

$$BP=\sqrt{(x+5)^2+(2x-5-5)^2}$$ $$=\sqrt{5(x^2-6x+25)}$$

Answer 2

Here is a solution with a geometric flavor:

You may know that the set of points $M$ such as $MA+MB=const.$ is an ellipse with focii $A$ and $B$. The larger the constant, the larger the ellipse. Consider the following figure:

If this constant is too small, the ellipse will not intersect the given line. There is a minimum value for which there is a unique point $M$ belonging to an ellipse of the family and to the straight line, and it is not difficult to see that in this point the line is tangent to the ellipse (colored in cyan).

A particular case occurs here: line $AB$ joining the focii is parallel to the given (green) line. Therefore the tangency will occur along the (black) line bissector of $AB$, with equation $y=-\tfrac12x$.

As a consequence, this minimum will occur at the intersection $P$ between the two lines, with coordinates:

$$P(x=2,y=-1)$$