General form $A x + B y + C = 0$

Every straight line in the plane can be written as $A x + B y + C = 0,$ where $A, B, C$ are real numbers and $A, B$ are not both zero. This general form is the canonical algebraic representation of a line. The art is to read off all geometric information , slope, intercepts, distance from origin, normal direction , directly from $A, B, C$.

Reading from the general form

For the line $A x + B y + C = 0$:

• Slope. Solve for $y$: $y = -\dfrac{A}{B} x - \dfrac{C}{B}$ (when $B \ne 0$). So $m = -\frac{A}{B}.$ If $B = 0$, the line is vertical: $x = -C/A$.

• $x$-intercept. Set $y = 0$: $x = -C/A$ (when $A \ne 0$).

• $y$-intercept. Set $x = 0$: $y = -C/B$ (when $B \ne 0$).

• Perpendicular distance from origin. $p = \dfrac{|C|}{\sqrt{A^2 + B^2}}$.

• Normal direction. The vector $(A, B)$ is perpendicular to the line. The unit normal is $\dfrac{1}{\sqrt{A^2 + B^2}}(A, B)$.

• Normal form. Divide $A x + B y + C = 0$ by $\pm\sqrt{A^2 + B^2}$ (choose sign to make the constant term negative , i.e., to get $p \ge 0$ on the right side): $\frac{A}{\sqrt{A^2 + B^2}} x + \frac{B}{\sqrt{A^2 + B^2}} y = \frac{-C}{\sqrt{A^2 + B^2}}.$

Two lines $A_1 x + B_1 y + C_1 = 0$ and $A_2 x + B_2 y + C_2 = 0$

• Parallel iff slopes equal iff $A_1 B_2 = A_2 B_1$ (and they are distinct: $A_1 C_2 \ne A_2 C_1$).
• Coincident iff $A_1/A_2 = B_1/B_2 = C_1/C_2$.
• Perpendicular iff $A_1 A_2 + B_1 B_2 = 0$.

Worked examples

Example 1. Find the slope and intercepts of $2 x - 3 y + 6 = 0$.

Slope: $m = -A/B = -2/(-3) = 2/3$. $x$-intercept: $-C/A = -6/2 = -3$. $y$-intercept: $-C/B = -6/(-3) = 2$.

Example 2. Perpendicular distance from origin to $3 x + 4 y - 10 = 0$.

$p = \dfrac{|-10|}{\sqrt{9 + 16}} = \dfrac{10}{5} = 2$.

Example 3. Express $4 x - 3 y + 10 = 0$ in normal form.

$\sqrt{A^2 + B^2} = 5$. Divide by $-5$ to make the right side positive: $-\dfrac{4}{5} x + \dfrac{3}{5} y = 2$. So $\cos\alpha = -4/5$, $\sin\alpha = 3/5$, $p = 2$.

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

Check: $A_1 B_2 = 2 \cdot (-6) = -12$, $A_2 B_1 = 4 \cdot (-3) = -12$. Equal. Are they coincident? $C_1/A_1 = 1/2, C_2/A_2 = -5/4$ , not equal. So parallel and distinct.

Example 5. Are $3 x - 4 y + 5 = 0$ and $8 x + 6 y - 1 = 0$ perpendicular?

$A_1 A_2 + B_1 B_2 = 3 \cdot 8 + (-4) \cdot 6 = 24 - 24 = 0$. Yes, perpendicular.

Try it yourself

1. Find the slope of $5 x + 2 y - 3 = 0$.
2. Convert $2 x + 3 y - 6 = 0$ to slope-intercept form.
3. Find the perpendicular distance from origin to $x + y - 4 = 0$.
4. Express $x - y + 5 = 0$ in normal form.
5. Are $x + 2 y - 4 = 0$ and $3 x + 6 y - 7 = 0$ parallel?
6. Are $5 x + 3 y - 1 = 0$ and $3 x - 5 y + 7 = 0$ perpendicular?
7. Find the equation of the line parallel to $3 x - 5 y + 7 = 0$ passing through $(1, -2)$.
8. Find the equation of the line perpendicular to $2 x + 3 y - 6 = 0$ passing through $(4, 5)$.
9. If $2 x + 3 y - 5 = 0$ and $5 x + p y - 3 = 0$ are parallel, find $p$.
10. The line $A x + B y + C = 0$ passes through the origin iff $C = ?$
11. Find the intercepts of the line $3 x - 2 y + 12 = 0$.
12. Convert $x + \sqrt{3} y - 6 = 0$ to normal form.

Pitfalls / Tricks

• If $B = 0$, the line is vertical and slope is undefined.
• For normal form, the choice of sign must give $p > 0$ on the right side.
• $(A, B)$ is normal to (perpendicular to) the line $A x + B y + C = 0$.
• Insight. The general form is just a relabelled "level set" of the linear function $f(x, y) = A x + B y + C$. The gradient $(A, B)$ points perpendicular to the line.