If I have three or more different points on a coordinate plane, is there a way to graph all the points where the sum of the distance to these three points is equal to a specific number.
For example, if I have point A (-5,0), point B (7,0) , and point C (1,4). How can I graph all the points whose distances to the original three points will sum up to a total of 14. I imagine it would look like semi circle above (1,0) but I don't know how to write it as a function.
On
If your $3$ points are $A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$, the point where you want to get the distance from is $(x,y)$ and the sum of the distances is $S$, then the function you're looking for is
$$\sqrt{(x - x_1)^2 + (y - y_1)^2} + \sqrt{(x - x_2)^2 + (y - y_2)^2} + \sqrt{(x - x_3)^2 + (y - y_3)^2} = S \tag{1}\label{eq1}$$
With your particular example of $A(-5,0), B(7,0), C(1,4)$, you would get, for them to add up to $14$, that
$$\sqrt{(x + 5)^2 + (y - 0)^2} + \sqrt{(x - 7)^2 + (y - 0)^2} + \sqrt{(x - 1)^2 + (y - 4)^2} = 14 \tag{2}\label{eq2}$$
With the $y_1$ and $y_2$ co-ordinates being $0$, this can be simplified to
$$\sqrt{(x + 5)^2 + y^2} + \sqrt{(x - 7)^2 + y^2} + \sqrt{(x - 1)^2 + (y - 4)^2} = 14 \tag{3}\label{eq3}$$
Let $P = (x,y)$ be a point satisfying this relationship.
The distance from $A$ to $P$ is $\sqrt{(x-(-5))^2 + (y-0)^2}$. If you can do the same for $B$ to $P$ and $C$ to $P$, your equation is just:
$$AP + BP + CP = 14$$
As it turns out, no point exists such that the sum of the distances is $14$. The minimum distance for a point to exist is around $14.3924766$ (Desmos graph).