Example — distance in a pencil of lines

A typical “shortcut” inequality: a distance to be compared with a value.

Example — Distance in a pencil of lines

In the pencil of lines $f_k:\ x-ky+k+2=0$, determine the values of $k$ for which the distance from the point $Q(1;2)$ to the line of the pencil is less than $1$.

Point–line distance: $d(Q,f_k)=\frac{|1\cdot 1 - k\cdot 2 + k+2|}{\sqrt{1+k^2}} = \frac{|3-k|}{\sqrt{1+k^2}}.$

Inequality: $\dfrac{|3-k|}{\sqrt{1+k^2}} < 1$.

Existence conditions: the denominator is $\sqrt{1+k^2}$ with $1+k^2>0$ always, so no restriction on $k$.

The denominator is strictly positive, so we can multiply both sides by $\sqrt{1+k^2}$ preserving the direction: $|3-k| < \sqrt{1+k^2}.$ We are in the case $|A| < \sqrt{B}$: both sides are $\ge 0$, we can square directly: $(3-k)^2 < 1+k^2 \iff 9-6k+k^2 < 1+k^2 \iff -6k < -8 \iff k>\tfrac{4}{3}.$

Solution: $\boxed{\,k>\tfrac{4}{3}\,}$.

Note how the absolute value and the root, both non-negative, allow squaring “with certainty”: the $|3-k|^2=(3-k)^2$ makes the modulus disappear without any sign study.

Links

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