Bisectors of two given lines

Example

Write the equations of the bisectors of the angles formed by $r:\ y - 2 = 0$ and $s:\ x - y = 0$.

Implicit form: $r:\ 0\cdot x + 1\cdot y - 2 = 0$, $s:\ 1\cdot x - 1\cdot y = 0$.

Locus condition: $\frac{|y_P - 2|}{\sqrt{0^2 + 1^2}} = \frac{|x_P - y_P|}{\sqrt{1^2 + (-1)^2}} \implies |y - 2| \cdot \sqrt{2} = |x - y|.$ Bringing $\sqrt{2}$ inside the first absolute value: $|\sqrt{2}(y-2)| = |x-y|$. We are in a case $|A| = |C|$, which is solved with the union of the two cases $A = C$ and $A = -C$: $\sqrt{2}(y-2) = x - y \ \cup\ \sqrt{2}(y-2) = -(x - y).$ Working through, we obtain: $b_1:\ y = (\sqrt{2}-1)x + 2(2-\sqrt{2}) \approx 0.414\,x + 1.17,$ $b_2:\ y = -(\sqrt{2}+1)x - 2(2+\sqrt{2}) \approx -2.414\,x - 6.83.$

Check. The two bisectors are perpendicular: $m_1 m_2 = (\sqrt{2}-1)(-\sqrt{2}-1) = -(\sqrt{2}^2 - 1) = -1$. ✓

The two lines $r$ and $s$ and their bisectors $b_1$ and $b_2$, mutually perpendicular.

Links

Topics: Loci
Concepts: Bisector · Point–line distance · Line · Absolute value
Skills: Analytic geometry · Solving equations