= Solution
Assume straight rays in a homogeneous medium. At horizontal distance $x-x_0$ the source distance is $R=\sqrt{(x-x_0)^2+z_0^2}$. With $v_A>v_B$,
$$
\Delta t=R\left(\frac1{v_B}-\frac1{v_A}\right),\qquad
(\Delta t)^2=M(x-x_0)^2+B,
$$
where $M=(1/v_B-1/v_A)^2$ and $B=(\Delta t_0)^2=M z_0^2$. Clock offsets cancel in a difference measured at one station. The difference curve is symmetric about the <epicentre> and reaches its minimum directly above the source.
The original plot uses the given numerical observations. Its broad minimum is consistent with
$$
\boxed{x_0\simeq140\pm15\,\mathrm{km},\qquad \Delta t_0\simeq15.0\pm0.5\,\mathrm{s}.}
$$
These are conservative graph-reading estimates, allowing for one-second rounding; the three equal smallest entries do not imply a flat physical minimum. A quadratic fit to squared differences refines the vertex to $x_0=140.17\,\mathrm{km}$ and $\Delta t_0=14.896\,\mathrm{s}$. These extra digits describe the fit, not the precision of an individual observation. Formal statistical errors require a model for the rounding and other measurement errors.
\Image[/past-exam-of-the-physics-of-the-earth-as-a-planet-course-of-the-university-of-cambridge/2012/paper-1-earthquake-location.png]
{title=Earthquake location fit}
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