Proving by using inequality of triangle

Question

suppose that points a and b are from different sides of a line m. Find a point y on line m such that the absolute difference of the YA and YB is maximal. Show proof.

Answer 1

From $\triangle AYB$, we have $ AB \ge \lvert YA - YB \rvert $.

Thus the maximum value would be $AB$, if we can show there can be equality for some $Y$. From the cosine law, $$AB^2 = YA^2+YB^2-2\cdot YA \cdot YB \cdot \cos \angle Y \to (YA-YB)^2$$ when $\angle Y \to 0$. Thus the value of $AB$ is never reached, but we can come arbitrarily close as we take $Y$ farther away from $A, B$ on either side of the line.