Various Forms of the Equation of a Line

The Equation of a Line: A Condition on Points

Every line in a plane contains infinitely many points. The central question is: given a line $L$ and an arbitrary point $P(x, y)$ in the $XY$-plane, how do we decide whether $P$ lies on $L$? The answer is that we need a condition — an algebraic equation in $x$ and $y$ — that is true exactly when $P$ is on $L$, and false otherwise. That equation is what we call the equation of the line.

The form this equation takes depends on what information we have about the line. Different pieces of data — a point and a slope, two points, intercepts, or the angle a perpendicular makes with the axes — lead to different, but equivalent, forms of the same equation. We now examine each of these forms in the order the textbook presents them.

---

1. Point-Slope Form

Suppose we know one specific point on the line and the slope of the line. Let the fixed point be $P_0(x_0, y_0)$ and let the slope be $m$. Take any other point $P(x, y)$ on the line. Since $P$ and $P_0$ both lie on the same line, the slope calculated from these two points must equal $m$:

Multiplying both sides by $(x - x_0)$ gives the point-slope form:

This is the equation of the line with slope $m$ passing through $(x_0, y_0)$.

Example. Find the equation of the line with slope $3$ passing through $(2, -5)$.

Here $m = 3$, $x_0 = 2$, $y_0 = -5$. Substituting:

---

2. Two-Point Form

If we know two distinct points on the line, say $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, we can first find the slope:

Then use the point-slope form with either point. Using $P_1$:

This is the two-point form of the equation of a line.

Example. Find the equation of the line through $(1, 2)$ and $(3, 6)$.

Slope $m = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$. Using point-slope with $(1, 2)$:

---

3. Slope-Intercept Form

Suppose we know the slope $m$ of a line and its $y$-intercept — the $y$-coordinate of the point where the line crosses the $y$-axis. Let the $y$-intercept be $c$, so the line passes through $(0, c)$. Using the point-slope form:

This simplifies to the slope-intercept form:

Here $m$ is the slope and $c$ is the $y$-intercept.

Example. Find the equation of a line with slope $-2$ and $y$-intercept $5$.

Directly: $y = -2x + 5$.

---

4. Intercept Form

Now suppose we know the $x$-intercept and $y$-intercept of a line. Let the line cut the $x$-axis at $A(a, 0)$ and the $y$-axis at $B(0, b)$, where $a \neq 0$ and $b \neq 0$. Using the two-point form with $A$ and $B$:

Simplify:

Multiply through by $a$:

Divide both sides by $ab$:

This is the intercept form of the equation of a line.

Example. Find the equation of a line whose $x$-intercept is $4$ and $y$-intercept is $-3$.

Here $a = 4$, $b = -3$. Substituting:

Multiply by $12$: $3x - 4y = 12$.

---

5. Normal Form

This form uses the perpendicular distance from the origin to the line and the angle that this perpendicular makes with the positive $x$-axis.

Let the line $L$ be at a perpendicular distance $p$ from the origin ($p > 0$). Let the perpendicular from the origin to $L$ meet $L$ at $N$, and let the angle that $ON$ makes with the positive $x$-axis be $\omega$ ($0 \leq \omega < 2\pi$).

We need to find the equation of $L$ in terms of $p$ and $\omega$.

Consider a point $P(x, y)$ on $L$. Draw perpendiculars from $P$ to the $x$-axis and to the line $ON$. The coordinates of $N$ are $(p\cos\omega, p\sin\omega)$.

›Proof

Derivation of the normal form

Let $L$ be the line, and let $ON$ be the perpendicular from the origin to $L$, with $ON = p$. The line $ON$ makes an angle $\omega$ with the positive $x$-axis, so the coordinates of $N$ are $(p\cos\omega, p\sin\omega)$.

The slope of $ON$ is $\tan\omega$. Since $L$ is perpendicular to $ON$, the slope of $L$ is $-\frac{1}{\tan\omega} = -\cot\omega$, provided $\tan\omega \neq 0$.

Using the point-slope form with point $N(p\cos\omega, p\sin\omega)$:

$y - p\sin\omega = -\cot\omega\,(x - p\cos\omega)$

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

Multiply both sides by $\sin\omega$:

$y\sin\omega - p\sin^2\omega = -\cos\omega\,(x - p\cos\omega)$

$y\sin\omega - p\sin^2\omega = -x\cos\omega + p\cos^2\omega$

Bring all terms to one side:

$x\cos\omega + y\sin\omega = p(\sin^2\omega + \cos^2\omega)$

Since $\sin^2\omega + \cos^2\omega = 1$, we obtain the normal form:

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

This is the equation of the line in normal form. Here $p$ is always taken as positive, and $\omega$ is the angle that the perpendicular from the origin makes with the positive $x$-axis.

Example. Find the equation of a line whose perpendicular distance from the origin is $5$ units and the perpendicular makes an angle of $60^\circ$ with the positive $x$-axis.

Here $p = 5$, $\omega = 60^\circ$. So $\cos 60^\circ = \frac{1}{2}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Substituting:

Multiply by $2$: $x + \sqrt{3}y = 10$.

--- …