How a Single Equation Describes Every Line Through One Point

Once a line has an equation, it can answer practical questions: how far a point lies from it, how far apart two parallel roads are, which lines pass through a crossing, and what path a moving point must follow. These ideas complete the study of straight lines.

This part covers normal and general forms, distances, families of lines and angle bisectors, and locus.

**A line whose perpendicular from the origin has length p and makes an angle $\alpha$ with the positive x-axis has normal form $x\cos \alpha +y\sin \alpha =p$, and every line can be written in general form $Ax+By+C=0$, with slope $-\frac{A}{B}$, x-intercept $-\frac{C}{A}$ and y-intercept $-\frac{C}{B}$.

General to normal form.** Divide by $\sqrt{A^2+B^2}$, choosing the sign that makes the right-hand side p positive.

Worked example. Reduce $\sqrt{3}x+y-8=0$ to normal form. Here $\sqrt{A^2+B^2}=\sqrt{3+1}=2$:

$$\frac{\sqrt{3}}{2}x+\frac{1}{2}y=4\Rightarrow x\cos 30°+y\sin 30°=4$$

The perpendicular from the origin has length $p=4$ and makes $30°$ with the x-axis.

An everyday example. A straight railway line on a city map can be described by how far it passes from the city centre and in which direction — exactly the information in normal form.

The substance. p is always positive — if dividing gives a negative right-hand side, divide by the negative root instead.

**The perpendicular distance of $(x_1,y_1)$ from $Ax+By+C=0$ is $d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$, and the distance between the parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is $\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.

Worked example.** Distance of $(3,-5)$ from $3x-4y-26=0$:

$$d=\frac{|3(3)-4(-5)-26|}{\sqrt{9+16}}=\frac{|9+20-26|}{5}=\frac{3}{5}=0.6$$

Worked example 2. Distance between $6x+8y-15=0$ and $3x+4y+5=0$. First match the coefficients by writing the second line as $6x+8y+10=0$. Then

$$d=\frac{|-15-10|}{\sqrt{36+64}}=\frac{25}{10}=2.5$$

An everyday example. Two parallel roads on a town plan, $3x+4y=12$ and $3x+4y=32$ with units of 100 m, are $\frac{20}{5}=4$ units, or 400 m, apart.

The substance. Make the x and y coefficients identical before using the parallel-lines formula — skipping this step is the commonest error.

**Every line through the intersection of $L_1=0$ and $L_2=0$, other than $L_2$ itself, can be written $L_1+k{L}_2=0$, and the bisectors of the angles between $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ are $\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}$.

Worked example (family).** Find the line through the intersection of $x+y-5=0$ and $2x-y-1=0$ that passes through the origin.

- Family: $(x+y-5)+k(2x-y-1)=0$
- At the origin: $-5-k=0$, so $k=-5$
- Line: $-9x+6y=0$, that is $3x-2y=0$

Check: the two lines meet at $(2,3)$, and $3(2)-2(3)=0$.

Worked example (bisectors). For $3x-4y+7=0$ and $12x+5y-2=0$:

$$\frac{3x-4y+7}{5}=\pm \frac{12x+5y-2}{13}$$

The plus sign gives $21x+77y-101=0$ and the minus sign gives $11x-3y+9=0$. Their slopes, $-\frac{3}{11}$ and $\frac{11}{3}$, multiply to $-1$, so the two bisectors are perpendicular.

An everyday example. All the straight roads that meet at one crossroads pass through a single point — a family of lines, one for each value of k.

The substance. Every point on a bisector is equidistant from both lines — that is exactly why the two distance expressions are set equal.

**A locus is the path traced by a point that moves according to a given condition, and its equation is found by taking the moving point as $(h,k)$, writing the condition algebraically, simplifying, and finally replacing h and k with x and y.

Steps:**

- Let the moving point be $P(h,k)$
- Translate the condition into an equation in h and k
- Simplify, then write x for h and y for k

Worked example. Find the locus of a point equidistant from $A(1,2)$ and $B(3,-4)$.

$$(h-1{)}^2+(k-2{)}^2=(h-3{)}^2+(k+4{)}^2$$

$$-2h-4k+5=-6h+8k+25\Rightarrow 4h-12k-20=0$$

The locus is $x-3y-5=0$, the perpendicular bisector of AB. Check: the midpoint $(2,-1)$ gives $2+3-5=0$.

Worked example 2. A point that stays 5 units from the origin satisfies $h^2+k^2=25$, so its locus is the circle $x^2+y^2=25$.

An everyday example. A goat tied to a peg by a 5 m rope and walking with the rope taut traces a circular locus around the peg.

The substance. The condition decides the shape — points equidistant from two lines, rather than two points, give the angle bisectors of the previous section.