The general equation and the normal form

One equation for every line

In the first course you wrote a line as $y = mx + c$, or by its intercepts. Each form misses some lines. A vertical line such as $x = 5$ has no slope. A line through the origin has no intercept form.

The general equation covers every line:

Multiply $A$, $B$ and $C$ by the same non-zero number and the line does not change. The equations $6x + 8y - 24 = 0$ and $3x + 4y - 12 = 0$ are the same line. Only the ratio $A : B : C$ matters, which is why two facts are enough to fix a line.

Reading the equation

Put $y = 0$ to find where the line cuts the $x$-axis. Put $x = 0$ to find where it cuts the $y$-axis. Solve for $y$ to read the slope.

• The $x$-intercept is $-\dfrac{C}{A}$, when $A \ne 0$.
• The $y$-intercept is $-\dfrac{C}{B}$, when $B \ne 0$.
• The slope is $-\dfrac{A}{B}$, when $B \ne 0$.

Example. For $3x + 4y - 12 = 0$, the slope is $-\dfrac{3}{4}$, the $x$-intercept is $4$ and the $y$-intercept is $3$.

Three cases need care.

• If $B = 0$, the line is vertical, $x = -\dfrac{C}{A}$, and it has no slope.
• If $A = 0$, the line is horizontal, $y = -\dfrac{C}{B}$, and its slope is $0$.
• If $C = 0$, the line passes through the origin, and both intercepts are $0$.

Parallel and perpendicular lines

Lines with the same $A$ and $B$ and different $C$ are parallel. So every line parallel to $Ax + By + C = 0$ has the form $Ax + By + k = 0$.

A line perpendicular to $Ax + By + C = 0$ has the form $Bx - Ay + k = 0$. The slopes are $-\dfrac{A}{B}$ and $\dfrac{B}{A}$, and their product is $-1$.

Two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ are

• parallel when $A_1B_2 = A_2B_1$,
• perpendicular when $A_1A_2 + B_1B_2 = 0$.

Example 1. Find the line through $(3, 2)$ parallel to $4x - 3y + 5 = 0$.

Step 1. Keep $A$ and $B$. The line is $4x - 3y + k = 0$.

Step 2. Use the point. $12 - 6 + k = 0$, so $k = -6$.

Answer: $4x - 3y - 6 = 0$.

Example 2. Find the line through $(2, 1)$ perpendicular to $2x + 5y - 7 = 0$.

Step 1. Swap and change one sign. The line is $5x - 2y + k = 0$.

Step 2. Use the point. $10 - 2 + k = 0$, so $k = -8$.

Answer: $5x - 2y - 8 = 0$.

Set $C = 0$ and watch the red segment shrink to nothing: a line through the origin is at distance $0$ from it.

The normal form

Drop the perpendicular from the origin to the line. Let its length be $p$, which is never negative. Let $\alpha$ be the angle it makes with the positive $x$-axis, measured from $0^\circ$ up to $360^\circ$.

The foot of the perpendicular is the point $(p\cos\alpha,\ p\sin\alpha)$. A point $(x, y)$ is on the line exactly when its shadow on the direction $(\cos\alpha, \sin\alpha)$ has length $p$. That gives the normal form:

Reducing $Ax + By + C = 0$ to the normal form

Step 1. Move the constant. Write $Ax + By = -C$.

Step 2. Make the right side positive. If $-C$ is negative, multiply the whole equation by $-1$.

Step 3. Divide by $\sqrt{A^2 + B^2}$. The coefficients of $x$ and $y$ become $\cos\alpha$ and $\sin\alpha$, and the right side is $p$.

Example 1. $3x + 4y - 12 = 0$. Write $3x + 4y = 12$. The right side is positive, and $\sqrt{9 + 16} = 5$. So

Here $\cos\alpha = 0.6$, $\sin\alpha = 0.8$ and $p = 2.4$. Both are positive, so $\alpha$ is in the first quadrant, and $\alpha = 53.13^\circ$.

Example 2. $x - y + 4 = 0$. Write $x - y = -4$. The right side is negative, so multiply by $-1$: $-x + y = 4$. Divide by $\sqrt{2}$:

Here $\cos\alpha = -\dfrac{1}{\sqrt{2}}$ and $\sin\alpha = \dfrac{1}{\sqrt{2}}$. The cosine is negative and the sine positive, so $\alpha = 135^\circ$. And $p = 2\sqrt{2} \approx 2.83$.

Example 3. The vertical line $x = 5$ is already in normal form with $\alpha = 0^\circ$ and $p = 5$. The horizontal line $y = -3$ becomes $-y = 3$, so $\alpha = 270^\circ$ and $p = 3$.

Always check the signs of $\cos\alpha$ and $\sin\alpha$ together. The value of $\tan\alpha$ alone cannot tell the second quadrant from the fourth.

Practise with new numbers

Try it in Python

The program reads the slope and the intercepts as exact fractions, finds the normal form, and checks a point of the line against it.

Common mistakes

• Reading the slope as $\dfrac{A}{B}$. Moving $Ax$ across changes its sign, so the slope is $-\dfrac{A}{B}$.
• Using the intercept form for a line through the origin. Both intercepts are $0$, and $\dfrac{x}{a} + \dfrac{y}{b} = 1$ cannot describe it.
• Looking for the slope of a vertical line. When $B = 0$ the line has no slope.
• Leaving the right side negative in the normal form. The number $p$ is a distance, so it must be positive.
• Taking $\alpha$ from $\tan\alpha$ alone. Use the signs of $\cos\alpha$ and $\sin\alpha$ to place the angle.

Remember these

• Every line is $Ax + By + C = 0$ with $A$ and $B$ not both $0$.
• Slope $-\dfrac{A}{B}$, $x$-intercept $-\dfrac{C}{A}$, $y$-intercept $-\dfrac{C}{B}$.
• Parallel to the line: $Ax + By + k = 0$. Perpendicular to the line: $Bx - Ay + k = 0$.
• Normal form: $x\cos\alpha + y\sin\alpha = p$, with $p \ge 0$.
• To reduce, write $Ax + By = -C$, make the right side positive, and divide by $\sqrt{A^2 + B^2}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the slope of the line $2x + 6y - 9 = 0$.

2. Find the intercepts of $5x - 4y + 20 = 0$.

3. Find the line through $(1, 4)$ parallel to $3x - y + 2 = 0$.

4. Are $2x + 3y - 5 = 0$ and $6x - 4y + 1 = 0$ perpendicular?

5. Find the distance $p$ of $5x + 12y - 26 = 0$ from the origin.