Notice that the third pair doen't even satisfy the constraint. Nonetheless,

... is by visualising, for sure.

Online 3-D Function Grapher

... and try to see the constraint as selecting just points on that surface (z = ...) that intersect with the plane (y = -1/2 x + 1) which is the plane passing vertically through the line y = -1/2 x + 1.

Then imagine getting the projection of that intersection against the x-z plane, where it will trace a curve which is z = just the result of replacing y with -1/2 x + 1. Sketch this curve, and locate the other two points to see that they are indeed critical.

PS: this problem may be for introducing you to Lagrange multipliers. Do the above first, though, and then plugging in the two equations as f and g here, Lagrange multiplier - Wikipedia, the free encyclopedia should make good sense.