Application — lines at an assigned distance

An irrational inequality (or equation) of the “shortcut” type appears very often when a geometric distance condition is translated into formulae. Let us look at a typical example that we shall meet again in the chapter on pencils of lines.

Example — Find the lines through $P$ at a fixed distance from $Q$

Determine the lines through $P(2;4)$ that are at distance $3$ from the point $Q(-2;4)$.

Set-up. A line through $P$ has equation $y-4=m(x-2)$, that is, in implicit form $mx - y - 2m + 4 = 0$. The distance of $Q(-2;4)$ from it is: $d(Q,r) = \frac{|m(-2) - 4 - 2m + 4|}{\sqrt{m^2+1}} = \frac{|-4m|}{\sqrt{m^2+1}} = \frac{4|m|}{\sqrt{m^2+1}}.$

Equation. $d(Q,r)=3$ becomes $\frac{4|m|}{\sqrt{m^2+1}} = 3 \iff 4|m| = 3\sqrt{m^2+1}.$

We are in a “shortcut” case: left-hand side $|A|$, right-hand side a positive multiple of $\sqrt{B}$, both $\ge 0$. We square directly without sign-splitting (the $|m|^2=m^2$ makes the absolute value disappear): $16 m^2 = 9(m^2+1) \iff 7m^2 = 9 \iff m^2 = \tfrac{9}{7} \iff m = \pm\tfrac{3}{\sqrt{7}}.$

The excluded case: the vertical line. The parametrisation $y-4=m(x-2)$ does not include the vertical lines. Let us check separately: the line $x=2$ is at distance $|-2-2|=4$ from $Q$, and $4\ne 3$, so it is not a solution.

Solution: there are exactly two lines, $\boxed{\,y-4 = \pm\tfrac{3}{\sqrt{7}}(x-2)\,}$.

Remark — Why we did not need to split

Had we ignored the structure and treated $4|m| = 3\sqrt{m^2+1}$ as “a root to isolate”, we would have added the conditions $4|m|\ge 0$ (always true) and $3\sqrt{m^2+1}\ge 0$ (always true). Neither of these restricts the domain: it is recognising the structure “both sides always $\ge 0$” that tells us, even before calculating, that such conditions will be useless.

Links

Topics: Irrational inequalities
Concepts: Irrational inequality · Point–line distance · Pencil of lines · Line · Absolute value
Methods: Squaring · Non-negative sides shortcut
Skills: Analytic geometry · Using formulae