Exercise $12.2.40$. If $r = \langle x,y \rangle$, $r_1 = \langle x_1, y_1\rangle$, $r_2 = \langle x_2, y_2\rangle$, describe the set of all points $(x,y)$ such that $|r - r_1| + |r + r_2| = k$ where $k > |r_1 - r_2|$.
I'm not sure how to do this. Here's my reasoning.
I look first at $|r - r_1| + |r + r_2| = k$ where $k = |r_1 - r_2|$. (A different problem.)
Given we're in two dimensions, the expression $|r - r_1| \leq k_1$ represents a disk of some radius centered at the tip of the vector $r_1$. (Similarly for $|r - r_2| \leq k_2$.) Also, the expression $|r_1 - r_2|$ represents the distance between the two disk centers.
Considering the expression $|r - r_1|$ it is just the distance from the tip of the vector $r$ to the tip of the vector $r_1$. So if I add this distance to $|r - r_2|$, I'm just getting a larger distance. If I want this larger distance to be equal to $k = |r_1 - r_2|$, then I think I get a circle of radius equal to the distance between the tips of $r_1$ and $r_2$.
The original problem asks for $k > |r_1 - r_2|$, which can be represented by any point outside the circle $|r - r_1| + |r - r_2| = |r_1 - r_2|$, hence giving me the complement of a disk of radius |r_1 - r_2|. (But I don't know where this disk is centered nor am I sure whether $|r - r_1| + |r - r_2|$ is a circle.)
So I don't know how to do this.
Reference: Calculus, James Stewart, 5th edition, exercise 12.2.40.
UPDATE $1$: After @OsamaGhani's hint, I managed to understand. The equation is the ellipse's equation in vector notation. So the equation represents the complement of an ellipsis with foci at the tip of $r_1$ and $r_2$. I've striked out my flawed original reasoning.
That $$ |r - r_1| + |r + r_2| = k$$ is the equation of an ellipse with foci located at $r_1$ and $-r_2$