Bisectors of two lines

Statement

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

Solution

Let us write the lines in implicit form: $r:\ 2x - y = 0$ and $s:\ x + y = 0$. A point $P(x;y)$ belongs to a bisector when $d(P,r) = d(P,s)$: $\frac{|2x - y|}{\sqrt{2^2 + (-1)^2}} = \frac{|x + y|}{\sqrt{1^2 + 1^2}} \implies \frac{|2x - y|}{\sqrt{5}} = \frac{|x + y|}{\sqrt{2}}.$ Hence $\sqrt{2}\,|2x - y| = \sqrt{5}\,|x + y|$, a case $|A| = |C|$ that splits into the two cases $A = C$ and $A = -C$: $\sqrt{2}(2x - y) = \sqrt{5}(x + y) \ \cup\ \sqrt{2}(2x - y) = -\sqrt{5}(x + y).$ The two bisectors are therefore: $b_1:\ (2\sqrt{2} - \sqrt{5})\,x - (\sqrt{2} + \sqrt{5})\,y = 0,\qquad b_2:\ (2\sqrt{2} + \sqrt{5})\,x + (\sqrt{5} - \sqrt{2})\,y = 0.$ In explicit form, with their gradients: $b_1:\ y = \frac{2\sqrt{2} - \sqrt{5}}{\sqrt{2} + \sqrt{5}}\,x \approx 0.162\,x,\qquad b_2:\ y = \frac{2\sqrt{2} + \sqrt{5}}{\sqrt{2} - \sqrt{5}}\,x \approx -6.16\,x.$ Check: $m_1 m_2 \approx 0.162 \cdot (-6.16) = -1$, so the two bisectors are perpendicular, as they must be. $\boxed{m_1 m_2 = -1}$

Links

Topics: Geometric loci
Concepts: Bisector · Point-line distance · Line · Absolute value
Skills: Analytical geometry · Solving equations
Exercise type: Geometric problem