Just think about this geometrically. As the slope of the line intersecting (2,2) increases, negatively away from m=-1, then the length of the segment AB increases without bound. Likewise, as the slope of the line increases, positively from m=-1, to zero m->0, the length of the segment AB increases without bound. It is natural to assume that the slope which is some where half way between the two, is our choice- a good guess being m=-1. To see this mathematically let us construct a function of the line passing through the point (2,2). So we have y=mx+b passing through (2,2) and wish to find m and b such that AB is minimized. Let us first use the fact that our line passes through (2,2) so that we may write m in terms of b. 2=2m+b implying m=(2-b)/b. Therefore the equation is now refined, y=((2-b)/b)x+b. After all we want to study the behavior as b changes, which is of course B by definition, since b=y-intercept. Now let us find A, in terms of b. Well A is the x-intercept, therefore it is when y=0, or 0=((2-b)/b)x+b which implies x=2b/(b-2)=A. Now we know the points A and B in terms of b in our equation. This allows us to change values of b to study the behavior of the length AB. Now we know the equation of the line and its endpoints A, B---in terms of b. So let us find the arclength of AB in terms of b. S(b)=int(((2-b)/b)x+b,0,2b/b-2), where we have integrated from the x-value of B and the x-value of A. We get S(b)=b^2/(b-2). Now we have the length of the segment AB in terms b. So, we can "play around" with values of the y-intercept, b, to see what would happen to the length of the segment. We can verify our early assumptions by the taking the following limits: lim[(b^2/(b-2)), b, inf]=inf and lim[b^2/(b-2), b, 2]=inf . Now to finish the problem we simply differentiate S(b) to find the critical points. S'(b)=[b(b-4)/(b-2)^2], implying that b=4 is our critical point giving us our minimum value for S(b). Plugging this value of b back into our equation, y=-x+4 is our desired equation.