Angle bisectors and line families

Two straight lines crossing on a page look like a settled, static picture — but hidden inside that single intersection point is an entire universe of lines. Every line through that point, no matter its slope, can be captured by one equation with a single dial you turn, called a parameter. And the two special lines that split the angles between the original pair exactly in half — the angle bisectors — can be pinned down without ever solving for the intersection point itself, using nothing more than the idea that every point on a bisector sits at equal distance from both lines. This section closes out ISC Class 11 Coordinate Geometry's Straight Lines unit by tying together slope, the two-line intersection, and the distance-from-a-line formula into two compact, exam-favourite techniques: the family of lines through an intersection, and the pair of angle-bisector equations.

The core insight

A line through the intersection of $L_1=0$ and $L_2=0$ can be written as $L_1+\lambda L_2=0$ for a parameter $\lambda$ , and the two angle bisectors between them are the loci where the distances from $L_1$ and $L_2$ are equal, obtained by setting $\dfrac{L_1}{\sqrt{A_1^2+B_1^2}} = \pm\dfrac{L_2}{\sqrt{A_2^2+B_2^2}}$ .

The core idea

Family of lines through an intersection. Suppose two lines $L_1: A_1x+B_1y+C_1=0$ and $L_2: A_2x+B_2y+C_2=0$ meet at a point. Instead of solving the pair of equations for that point's coordinates, notice that every line through it shares a common structure.

A_1x+B_1y+C_1 + \lambda(A_2x+B_2y+C_2) = 0

where: $\lambda$ is a free real-number parameter, and the expression on the left is itself the equation of a straight line for every fixed value of $\lambda$ . As $\lambda$ ranges over all real numbers, this equation sweeps out every line through the shared intersection point except $L_2$ itself (that member would need $\lambda \to \infty$ ). Once an extra condition is given — the line must pass through a named third point, have a stated slope, or be parallel/perpendicular to something — substituting that condition into the combined equation pins down a unique $\lambda$ , and hence the unique required line, with no need to ever compute the intersection point's coordinates.

Angle bisectors of two lines. A bisector of the angle between two lines is the set of points equally distant from both lines. Using the perpendicular-distance-from-a-line formula for a general point $(x,y)$ and setting the two distances equal, with a $\pm$ sign to allow for either of the two bisectors, gives the pair

\frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}} = \pm\, \frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}

where: the $+$ sign traces out one bisector and the $-$ sign the other. These two bisector lines always meet each other at a right angle, because together they must split the full $360°$ around the intersection point into four equal right-angle-adjacent regions — one bisector serving the acute angle between $L_1$ and $L_2$ , the other the obtuse angle.

Telling the acute bisector from the obtuse one. First rewrite both line equations so the constant terms $C_1$ and $C_2$ carry the same sign. With that done, check the sign of $A_1A_2+B_1B_2$ : if it is negative, the $+$ sign in the bisector formula gives the acute-angle bisector; if it is positive, the $+$ sign instead gives the obtuse-angle bisector (so the $-$ sign gives the acute one). This test works because $A_1A_2+B_1B_2$ is proportional to the cosine of the angle between the two lines' normal directions once the constants are sign-matched.

Figure

Two intersecting lines with the acute angle and obtuse angle marked, and the two perpendicular bisector lines drawn through the same intersection point, one splitting the acute angle and one splitting the obtuse angle

Derivation ADeriving the two angle-bisector equations from the equal-distance condition

Starting from the perpendicular-distance formula for a point from a line, show why a point $(x,y)$ lying on a bisector of the angle between $L_1: A_1x+B_1y+C_1=0$ and $L_2: A_2x+B_2y+C_2=0$ must satisfy $\dfrac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}} = \pm\dfrac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}$ , and why the two choices of sign hand you two mutually perpendicular lines rather than one.

Step 1. Recall that the perpendicular distance of a point $(x,y)$ from a line $Ax+By+C=0$ is $\dfrac{|Ax+By+C|}{\sqrt{A^2+B^2}}$ . A point lies on the bisector of the angle between $L_1$ and $L_2$ precisely when it is equally far from both lines, so its distance from $L_1$ must equal its distance from $L_2$ .

Step 2. Write that equality of distances as $\dfrac{|A_1x+B_1y+C_1|}{\sqrt{A_1^2+B_1^2}} = \dfrac{|A_2x+B_2y+C_2|}{\sqrt{A_2^2+B_2^2}}$ . Because both sides carry absolute-value bars, the underlying algebraic quantities inside them can match in magnitude while differing in sign — the absolute-value equation is satisfied whenever the two unsigned expressions are equal, regardless of whether the signed expressions themselves are equal or negatives of each other.

Step 3. Removing the absolute-value bars therefore produces two separate algebraic cases rather than one: either the two signed expressions are equal, or one is the negative of the other. Writing both cases together with a single $\pm$ sign gives $\dfrac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}} = \pm\, \dfrac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}$ , which is a pair of two distinct linear equations in $x$ and $y$ — one line for the $+$ choice and a different line for the $-$ choice.

Step 4. Both resulting lines pass through the same intersection point of $L_1$ and $L_2$ , since at that point both original distances are zero and the equal-distance condition is trivially satisfied for either sign. Around that shared point, the two original lines create two pairs of vertically opposite angles — one acute pair and one obtuse pair — and a bisector line must split one of these pairs exactly in half. Since the acute-angle bisector and the obtuse-angle bisector split angles that together add to $90°$ (half of the acute angle plus half of its supplementary obtuse angle equals $45°+45°=90°$ ), the two bisector lines obtained from the $+$ and $-$ cases are always perpendicular to each other, completing the derivation.

Worked example

Find the equations of the bisectors of the angles between the lines $3x-4y+7=0$ and $12x-5y-8=0$ , and state which bisector splits the acute angle.

Step 1. Identify the coefficients: $A_1=3,\,B_1=-4,\,C_1=7$ and $A_2=12,\,B_2=-5,\,C_2=-8$ . Compute the normalising square roots: $\sqrt{A_1^2+B_1^2}=\sqrt{9+16}=5$ and $\sqrt{A_2^2+B_2^2}=\sqrt{144+25}=13$ .

Step 2. Set up the equal-distance equation with the $\pm$ sign: $\dfrac{3x-4y+7}{5} = \pm\,\dfrac{12x-5y-8}{13}$ .

Step 3. Taking the $+$ sign and cross-multiplying: $13(3x-4y+7)=5(12x-5y-8)$ , so $39x-52y+91 = 60x-25y-40$ , which rearranges to $21x+27y-131=0$ .

Step 4. Taking the $-$ sign and cross-multiplying: $13(3x-4y+7)=-5(12x-5y-8)$ , so $39x-52y+91=-60x+25y+40$ , which rearranges to $99x-77y+51=0$ .

Step 5. To decide which bisector is acute, first check that $C_1=7$ and $C_2=-8$ have opposite signs, so rewrite $L_2$ as $-12x+5y+8=0$ , giving $C_2'=8$ , now matching the sign of $C_1$ . Recompute $A_1A_2'+B_1B_2' = 3(-12)+(-4)(5) = -36-20=-56$ , which is negative, so by the sign rule the $+$ case bisector is the acute-angle bisector. The two bisectors are $21x+27y-131=0$ and $99x-77y+51=0$ , and $21x+27y-131=0$ is the acute-angle bisector.

Competitive stretch

The same equal-distance condition can be applied without first finding the family-of-lines equation: for the line through $(1,2)$ and the intersection of $x+y-3=0$ and $2x-y+1=0$ , substituting the family equation $(x+y-3)+\lambda(2x-y+1)=0$ at $(1,2)$ gives $0+\lambda(1)=0$ , so $\lambda=0$ and the required line is simply $x+y-3=0$ — because $(1,2)$ already lies on that line, no intersection point ever needed to be computed.

Common traps

Forgetting that no finite $\lambda$ in the family-of-lines equation recovers $L_2$ itself — that case must be checked separately if the target line looks like it could be $L_2$ .
Substituting the extra condition into $L_1$ or $L_2$ alone instead of into the combined family equation $L_1+\lambda L_2=0$ .
Applying the $A_1A_2+B_1B_2$ sign test for the acute bisector before matching the signs of $C_1$ and $C_2$ , which silently flips the answer.
Sign or arithmetic slips while cross-multiplying by the two square-root normalising factors, since several terms must be expanded correctly on both sides.

Ask the tutor: If the same-sign-of-constants step before picking the acute bisector still feels shaky, ask the AI tutor to walk through it with a fresh pair of lines you pick yourself.

Why does substituting a point into the family-of-lines equation work even without knowing the intersection point

The family equation $L_1+\lambda L_2=0$ is built so that at the actual intersection point, both $L_1$ and $L_2$ evaluate to zero, so the combined expression is zero for every $\lambda$ automatically — the intersection point never has to be located. Substituting any other named point simply gives one linear equation in the single unknown $\lambda$ , which is enough to fix it.

What if both signs in the bisector formula give equations that look equally "nice" — how do I know which is genuinely the acute one without redoing the whole sign check

There is no shortcut around the sign check: you must first make $C_1$ and $C_2$ carry the same sign (multiplying one whole equation by $-1$ if needed), then test the sign of $A_1A_2+B_1B_2$ . A negative value marks the $+$ case as acute; a positive value marks the $+$ case as obtuse. Skipping the sign-matching step is the single most common way students get this backwards.

Is the family-of-lines idea only useful for two straight lines, or does it appear elsewhere too

Within this syllabus it is used specifically for two straight lines, but the same logic — that $S_1+\lambda S_2=0$ passes through the intersection of curves $S_1=0$ and $S_2=0$ — reappears later with circles and other conics, so mastering the reasoning here (not just the line version) pays off in later topics.

Angle bisectors and line families

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads