Using the spacetime diagram, calculate the speed $v$ of the $Σ'$ frame relative to the $Σ$ frame, in terms of $t_A, t_B, c, x_A, x_B$

As measured in frame $$Σ$$, the coordinates of events $$A$$ and $$B$$ are respectively $$(ct_A,x_A)$$ and $$(ct_B,x_B)$$. The interval between the events are spacelike. There is then a frame $$Σ'$$ in which the two events are simultaneous.

Using the spacetime diagram, calculate the speed $$v$$ of the $$Σ'$$ frame relative to the $$Σ$$ frame, in terms of $$t_A, t_B, c, x_A, x_B$$.

Using a Lorentz transformation I found quite easily that: $$\Delta t'=\gamma_v (\Delta t - \frac{v\Delta x}{c^2})$$ which reduces to:

$$v=(\frac{t_A-t_B}{x_A-x_B})c^2$$ as $$\Delta t'=0$$.

I am struggling to derive this equation from the spacetime diagram. I have constructed a diagram similar to the one seen here {https://physics.stackexchange.com/q/367150} but I don't fully understand how to complete this problem from this.

I know I am just missing something simple but I could do with some help. Sorry for the terrible graph but that's all I can do. For simplicity let me take A as the origin of the $$\Sigma$$ such that $$A(ct_A, x_A)$$ and $$B(ct_B, x_B)$$.
$$A$$ and $$B$$ are simultaneous means that they lie on the same plane w.r.t $$\Sigma'$$. Which mathematically also corresponds to $$t'=0$$. As you can see there is an angle between them so by using the below equation, we can obtain $$tan(\alpha) = v/c$$
$$\frac{c(t_B - t_A)}{x_B-x_A} = \frac{v}{c}$$