Normal, symmetric and general forms

Every straight line can be writ­ten as $ax + by + c = 0$, the gen­eral form. This les­son shows how to read the slope and inter­cepts from it and how to con­vert it into the other forms. It also intro­duces two more forms: the nor­mal form, which describes a line by its dis­tance from the ori­gin and the direc­tion of the per­pen­dic­u­lar to it, and the sym­met­ric (dis­tance) form, which locates the points on a line at a given dis­tance from a fixed point.

Nor­mal form#

Draw the per­pen­dic­u­lar from the ori­gin $O$ to the line, meet­ing it at $N$. Let $p = ON$ be the length of this per­pen­dic­u­lar ($p > 0$), and let $\alpha$ be the angle that $ON$ makes with the pos­i­tive $x$-axis, mea­sured anti­clock­wise ($0° \leq \alpha < 360°$).

Result. The equa­tion of the line is

$$x\cos\alpha + y\sin\alpha = p$$

Why. The point $N$ is $(p\cos\alpha, p\sin\alpha)$, and the slope of $ON$ is $\tan\alpha$. The line is per­pen­dic­u­lar to $ON$, so its slope is $\displaystyle -\frac{\cos\alpha}{\sin\alpha}$. Through $N$:

$$\displaystyle y - p\sin\alpha = -\frac{\cos\alpha}{\sin\alpha}(x - p\cos\alpha)$$

Mul­ti­ply­ing by $\sin\alpha$ and using $\sin^2\alpha + \cos^2\alpha = 1$ gives $x\cos\alpha + y\sin\alpha = p$. (The cases $\sin\alpha = 0$ give ver­ti­cal lines and can be checked directly.)

Exam­ple 1#

Find the line whose per­pen­dic­u­lar dis­tance from the ori­gin is $4$, the per­pen­dic­u­lar mak­ing $30°$ with the pos­i­tive $x$-axis.

$$\displaystyle x\cos 30° + y\sin 30° = 4 \implies \frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4 \implies \sqrt{3}x + y = 8$$

Gen­eral form#

Every line has an equa­tion of the form

$$ax + by + c = 0, \qquad a \text{ and } b \text{ not both zero}$$

and every such equa­tion is a straight line. From the gen­eral form:

• Slope $\displaystyle = -\frac{a}{b}$ (if $b \neq 0$; if $b = 0$ the line is ver­ti­cal).
• $x$-inter­cept $\displaystyle = -\frac{c}{a}$ and $y$-inter­cept $\displaystyle = -\frac{c}{b}$ (when $a$, $b$ are non-zero).

Exam­ple 2#

For $2x - 3y + 12 = 0$, find the slope and inter­cepts, and write the line in slope-inter­cept and inter­cept forms.

Slope $\displaystyle = -\frac{2}{-3} = \frac{2}{3}$. The $x$-inter­cept is $\displaystyle -\frac{12}{2} = -6$, and the $y$-inter­cept is $\displaystyle -\frac{12}{-3} = 4$.

Slope-inter­cept form: $3y = 2x + 12$, so $\displaystyle y = \frac{2}{3}x + 4$.

Inter­cept form: $2x - 3y = -12$; divid­ing by $-12$, $\displaystyle \frac{x}{-6} + \frac{y}{4} = 1$.

Con­vert­ing to nor­mal form#

To bring $ax + by + c = 0$ into nor­mal form:

1. Move the con­stant to the right: $ax + by = -c$. If $-c$ is neg­a­tive, mul­ti­ply through by $-1$ so the right side is pos­i­tive (the right side must be $p > 0$).
2. Divide through by $\sqrt{a^2 + b^2}$. The coef­fi­cients of $x$ and $y$ are then $\cos\alpha$ and $\sin\alpha$, and the right side is $p$.

Exam­ple 3#

Write $x + \sqrt{3}y - 6 = 0$ in nor­mal form, and find $p$ and $\alpha$.

$x + \sqrt{3}y = 6$, and $\sqrt{1 + 3} = 2$. Divid­ing by $2$:

$$\displaystyle \frac{1}{2}x + \frac{\sqrt{3}}{2}y = 3$$

So $\displaystyle \cos\alpha = \frac{1}{2}$, $\displaystyle \sin\alpha = \frac{\sqrt{3}}{2}$, giv­ing $\alpha = 60°$, and $p = 3$.

Exam­ple 4#

Write $\sqrt{3}x - y + 4 = 0$ in nor­mal form.

$\sqrt{3}x - y = -4$. The right side is neg­a­tive, so mul­ti­ply by $-1$: $-\sqrt{3}x + y = 4$. Divide by $\sqrt{3 + 1} = 2$:

$$\displaystyle -\frac{\sqrt{3}}{2}x + \frac{1}{2}y = 2$$

So $\displaystyle \cos\alpha = -\frac{\sqrt{3}}{2}$ and $\displaystyle \sin\alpha = \frac{1}{2}$: $\alpha$ is in the sec­ond quad­rant, $\alpha = 150°$, and $p = 2$.

Sym­met­ric (dis­tance) form#

Result. The line through $(x_1, y_1)$ with incli­na­tion $\theta$ can be writ­ten

$$\displaystyle \frac{x - x_1}{\cos\theta} = \frac{y - y_1}{\sin\theta} = r$$

where $r$ is the dis­tance from $(x_1, y_1)$ to the point $(x, y)$, mea­sured along the line (pos­i­tive in the direc­tion of increas­ing $x$ when $\theta$ is acute). So the point at dis­tance $r$ along the line is

$$(x_1 + r\cos\theta, \; y_1 + r\sin\theta)$$

Exam­ple 5#

Find the point at a dis­tance $2$ from $(1, 2)$ along the line of incli­na­tion $60°$, in the direc­tion of increas­ing $x$.

$$(1 + 2\cos 60°, \; 2 + 2\sin 60°) = (1 + 1, \; 2 + \sqrt{3}) = (2, 2 + \sqrt{3})$$

The point at dis­tance $2$ on the other side is $(1 - 1, 2 - \sqrt{3}) = (0, 2 - \sqrt{3})$.

Prac­tice#

1. Find the line in nor­mal form with $p = 5$ and $\alpha = 60°$, and write it in gen­eral form.
2. Find the line with $p = 3$ and $\alpha = 135°$, in gen­eral form.
3. Write $x + y - 4 = 0$ in nor­mal form, and find $p$ and $\alpha$.
4. Write $\sqrt{3}x + y + 6 = 0$ in nor­mal form, and find $p$ and $\alpha$.
5. Write $3x + 4y - 15 = 0$ in nor­mal form, and find $p$, $\cos\alpha$ and $\sin\alpha$.
6. Find the slope and inter­cepts of $4x + 5y - 20 = 0$ from the gen­eral-form for­mu­lae.
7. Write $5x - 2y - 10 = 0$ in slope-inter­cept form and in inter­cept form.
8. Find the point at dis­tance $4$ from $(2, 1)$ along the line of incli­na­tion $30°$, in the direc­tion of increas­ing $x$.
9. Write the line $y = 4$ in nor­mal form.
10. Write the line $x = -3$ in nor­mal form.

Answers#