任意等距变换中如果能确定一条直线上两点，那么就可以确定直线所有的点

Suppose that we have an isometry $F$ with two fixed points $A$ and $B$. Thus, $F(A)=A$ and $F(B)=B$.

Consider a point $X$ on the line $AB$. Let $XA=a$ and $XB=b$. Let $Y=F(X)$.

Since $F$ is an isometry, $Y$ must lie on the hypersphere (the number of dimensions will remain unspecified for maximal generality) with center $A$ and radius $a$.

Similarly, $Y$ must also lie on the hypersphere with center $B$ and radius $b$.

But wait! Since $X$ lies on the line $AB$, one of the quantities ${a+b, a-b, b-a}$ must equal $AB$!

This means that the two hyperspheres mentioned above are tangent, and what is their single point of intersection?

You guessed it; it's our point $X$! Clearly, we know from the conditions that $X$ satisfies the two conditions; moreover, since we know that the hyperspheres are tangent, no other point could work.

Thus, we have that $Y=F(X)=X$, which means that $X$ is also a fixed point of the isometry.

Since $X$ was an arbitrary point on the line $AB$, the same must hold for all points on the line; thus, we have that any isometry that fixes two points fixes all points on the line containing them.