Answer: This is geometry. Put point C at (-4,-3). Distance from A to C = AC = 5. Distance from C to B = CB = 9. Then ACB is a right triangle with legs AC and CB, and hypotenuse AB, so
AC^2 + CB^2 = AB^2, so
AB = √(AC^2 + CB^2) = √(5^2+9^2) = √(25+81) = √106.
However, in vector algebra, A=(-4,3) and B=(5,-2), and (A-B)=(-9,5).
distance(A,B) = √((A-B) dot (A-B)),
where (x,y) dot (x,y) = x×x + y×y.
So the answer is √((-9,5)dot(-9,5)) = √((-9)^2+5^2) = √(81+25) = √106.
This works in 3D, (x,y,z) dot (x,y,z) = x×x + y×y + z×z. distance(A,B) gives √(x^2+y^2+z^2) for distances in 3 dimensions.
In 4D, distance is √(x^2+y^2+z^2+w^2)
And in infinite dimensional Hilbert space, (x,y,z,a,b,c,...) dot (x,y,z,a,b,c,...) = x×x + ... c×c + .... distance(A,B) gives
√(x^2+...+c^2+...).
And for real valued functions, f(x) dot g(x) is roughly the sum of f(x)×g(x) over uncountably many points x from -infinity to +infinity. It's the area under h(x)=f(x)g(x).