Example — Fixed distance

When the condition concerns the distance from a point, one applies the point–line distance formula and obtains an inequality (or equation) in the parameter $k$.

Example — Fixed distance

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

Distance. We write the line in implicit form $1\cdot x + (-k)\cdot y + (k+2) = 0$ and apply the point–line distance formula: $d(Q, f_k) = \frac{|1\cdot 1 - k\cdot 2 + k+2|}{\sqrt{1^2 + (-k)^2}} = \frac{|3-k|}{\sqrt{1+k^2}}.$

Inequality. $\dfrac{|3-k|}{\sqrt{1+k^2}} < 1$. The denominator is strictly positive, so it can be moved to the numerator: $|3-k| < \sqrt{1+k^2}.$ Both sides are $\ge 0$, so we are in the “shortcut” case of the previous chapter: we square directly. $(3-k)^2 < 1+k^2 \iff 9-6k+k^2 < 1+k^2 \iff k > \tfrac{4}{3}.$

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

Links

Topics: Pencils of lines
Concepts: Point–line distance · Implicit equation · Pencil of lines · Pencil parameter · Line
Functions: Line
Skills: Analytic geometry · Solving inequalities · Using formulae