Slope of a line

A straight line, extended in both directions, makes some angle $\theta$ with the positive direction of the x-axis, always measured going counter-clockwise from that axis. By convention $\theta$ is taken in the range $0^\circ \le \theta \le 180^\circ$, since going further would just repeat the same line. The slope of the line is defined as the tangent of this angle:

where: $m$ is the slope (a pure number, no units), and $\theta$ is the inclination of the line with the positive x-axis. The single formula splits cleanly into four cases that are worth memorising as a set, because most exam questions are really just testing whether you can read $\theta$ off a picture and place it in the right case.

• $\theta = 0^\circ$: the line is horizontal, and $m = \tan 0^\circ = 0$.
• $0^\circ < \theta < 90^\circ$: the line rises from left to right, and $m > 0$.
• $90^\circ < \theta < 180^\circ$: the line falls from left to right, and $m < 0$.
• $\theta = 90^\circ$: the line is vertical; $\tan 90^\circ$ is not defined, so a vertical line has no slope at all — not zero, not infinite, simply undefined.

Slope belongs to the line itself, not to any particular pair of points chosen on it — two different points on the same line always give the same $m$ (this equivalence is proved once coordinates are introduced, in the companion topic on computing slope from two points). The picture below fixes the four cases in your mind before you meet a single numeric example.

The everyday meaning of "steepness" is rise over run: how much a path climbs vertically for every unit it moves horizontally. If you draw a right triangle with the line as its hypotenuse, the vertical rise sits opposite the angle $\theta$ at the base, and the horizontal run sits adjacent to it, so trigonometry gives rise over run exactly as $\tan\theta$ — which is why $m = \tan\theta$ matches the intuitive rise-over-run picture used for ramps and roofs, and why mathematicians simply promoted that everyday ratio into a formal definition tied to the angle $\theta$ measured from the positive x-axis so that the sign of $m$ automatically records the direction of tilt as well as its steepness. As $\theta$ grows from $0^\circ$ towards $90^\circ$, $\tan\theta$ grows without bound, so lines get steeper and steeper the closer they swing towards vertical; the moment $\theta$ actually reaches $90^\circ$ there is no finite rise-over-run left to state, which is exactly why the vertical line's slope is undefined rather than some very large number. Past $90^\circ$, the rise turns negative relative to the run, flipping $\tan\theta$ negative and matching a line that now falls as $x$ increases — the sign of $m$ is therefore not a bookkeeping trick but a direct geometric consequence of which way the triangle's rise points. This also explains why the definition insists $0^\circ \le \theta \le 180^\circ$ rather than allowing any angle at all: a straight line, unlike a ray, extends equally in two opposite directions, and the angle measured going one way along it is simply $180^\circ$ minus the angle measured going the other way. Restricting $\theta$ to a single half-turn picks out one number per line instead of two candidates that would otherwise disagree. One more relation worth keeping alongside the definition, ready for the next companion topic (parallelism and perpendicularity): the acute angle $\theta$ between two lines of slopes $m_1$ and $m_2$ satisfies

where: $m_1, m_2$ are the slopes of the two lines and $\theta$ is the acute angle between them.

Problem. A line makes an angle of $135^\circ$ with the positive direction of the x-axis. Find its slope and describe whether the line rises or falls as $x$ increases. Then, separately, a ramp rises $1$ metre in height for every $5$ metres of horizontal run — state its slope and the sign of its inclination angle.

Step 1. For the $135^\circ$ line, apply the definition directly: $m = \tan 135^\circ$. Write $135^\circ = 180^\circ - 45^\circ$ and use $\tan(180^\circ - \alpha) = -\tan\alpha$, so $\tan 135^\circ = -\tan 45^\circ = -1$.

Step 2. Interpret the sign: since $m = -1 < 0$ and $90^\circ < 135^\circ < 180^\circ$, this matches the third case above exactly, so the line falls as $x$ increases.

Step 3. For the ramp, the rise-over-run description is itself the slope: $m = \dfrac{\text{rise}}{\text{run}} = \dfrac{1}{5} = 0.2$. Because the ramp climbs going forward, its inclination $\theta$ satisfies $0^\circ < \theta < 90^\circ$, consistent with $m = 0.2 > 0$.

Step 4. To pin down $\theta$ itself for the ramp we would write $\theta = \tan^{-1}(0.2)$, but inverse trigonometric functions are outside the Class 11 syllabus — it is enough here to state that $\theta$ is some acute angle satisfying $\tan\theta = 0.2$, without evaluating it numerically.

Answer: for the $135^\circ$ line, $m = -1$, falling left to right; for the ramp, $m = 0.2$, an acute rising angle with $\tan\theta = 0.2$.

• Treating a vertical line as having "zero slope" — it is the horizontal line that has $m = 0$; a vertical line has no defined slope at all.
• Measuring $\theta$ from the wrong axis or in the wrong direction — it must always be from the positive x-axis, counter-clockwise, and kept within $0^\circ$ to $180^\circ$.
• Assuming steepness increases smoothly and monotonically across the whole $0^\circ$ to $180^\circ$ sweep — in fact $|\tan\theta|$ grows as $\theta \to 90^\circ$ from either side and then shrinks again past it, so equal jumps in $\theta$ do not mean equal jumps in steepness.
• Reading a negative slope as meaning the line is somehow "smaller" — it only records that the line falls as $x$ increases, not anything about its length or magnitude.

Ask the tutor: if you are still unsure why a vertical line "has no slope" instead of an infinite one, or want to practice reading $\theta$ off a diagram, ask your AI tutor to walk through a few more slope examples with you.