Finding a simple fractional expression of a bilinear transformation given two fixed points.

If $z_1$ and $z_2$ are distinct fixed points of a bilinear transformation $w=T(z)$ show that the transformation may be expressed as

$$\frac{w-z_1}{w-z_2}=K\frac{z-z_1}{z-z_2},$$

where $K$ is a complex constant.

I'm trying to use this fact to solve the problem.

So we have $w_2=z_2, w_1=z_1$, in this case, however, plugging these values in, I cannot get the desired expression, so I'm stuck here. How can I get this expression? I would greatly appreciate any help.

If the Möbiustransformation $T$ maps $z_1, z_2, z_3 \in \Bbb C$ to $w_1, w_2, w_3 \in \Bbb C$ then $$\frac{(T(z)-w_1)(w_3 - w_2)}{(T(z)-w_2)(w_3-w_1)} = \frac{(z-z_1)(z_3 - z_2)}{(z-z_2)(z_3-z_1)}$$ (Note that you can arrange the three points – together with their image points – in any convenient order.)
If $z_1$ and $z_2$ are (finite) fixed points of $T$ then choose any $z_3 \in \Bbb C$ such that $w_3 = T(z_3) \in \Bbb C$, and you obtain $$\frac{(T(z)-z_1)(w_3 - z_2)}{(T(z)-z_2)(w_3-z_1)} = \frac{(z-z_1)(z_3 - z_2)}{(z-z_2)(z_3-z_1)}$$ which is the desired representation.
Another approach: Define $$S(z) = \frac{z-z_1}{z-z_2} \, .$$ Then $S \circ T \circ S^{-1}$ has the fixed points $0$ and $\infty$. Setting $K := S \circ T \circ S^{-1}(1)$ it follows that $$S \circ T \circ S^{-1}(w) = Kw$$ because the left-hand side and the right-hand side are Möbiustransformations mapping $0, 1, \infty$ to $0, K, \infty$. Therefore $$S(T(z)) = K S(z)$$ which is again the desired representation.