Slope of a Line

Inclination and Slope

A line crossing the $x$-axis makes two supplementary angles with it. The angle $\theta$ measured anticlockwise from the positive $x$-axis is the inclination of the line, with $0° \le \theta < 180°$.

Key Point (Definition): The slope (or gradient) of a line with inclination $\theta$ is $m = \tan\theta$, $\theta \ne 90°$. The $x$-axis has slope 0; a vertical line has no slope — $\tan 90°$ is undefined.

Key Point (Two-point formula): The line through $P(x_1, y_1)$ and $Q(x_2, y_2)$ with $x_1 \ne x_2$ has

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

The textbook proof runs both the acute and obtuse cases — the same formula emerges, so you never need to know which case you are in.

[Board Tip] Four little computations are the whole story: $(3,-2)$ to $(-1,4)$ gives $m = -\frac{3}{2}$ (falling); equal $y$'s give $m = 0$ (horizontal); equal $x$'s give an undefined slope (vertical); inclination $60°$ gives $m = \sqrt{3}$.

Parallel, Perpendicular, and the Angle Between Lines

Key Point: For non-vertical lines: parallel $\Leftrightarrow m_1 = m_2$ (equal inclinations) and perpendicular $\Leftrightarrow m_1 m_2 = -1$ (slopes are negative reciprocals). Three points $A, B, C$ are collinear exactly when slope of $AB$ = slope of $BC$.

When two lines cross, they make a pair of supplementary angles. In terms of slopes:

Key Point (Angle formula): The acute angle $\theta$ between lines with slopes $m_1, m_2$ satisfies

$$\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right|, \qquad 1 + m_1m_2 \ne 0$$

The obtuse angle is $180° - \theta$.

For example: a line makes $45°$ with a line of slope $\frac{1}{2}$ — solving $\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = \pm 1$ gives two slopes, $m = 3$ or $m = -\frac{1}{3}$: one line on each side of the given one.

[JEE Tip] The $\pm$ is not a nuisance — it is geometry. Any "line making angle $\alpha$ with a given line" problem has two answers unless a further condition (a point, a quadrant) kills one.

Solved Examples

Example 1: Slopes four ways

Find the slope of the line (a) through $(3, -2)$ and $(-1, 4)$ (b) through $(3, -2)$ and $(7, -2)$ (c) through $(3, -2)$ and $(3, 4)$ (d) with inclination $60°$.

Solution:

Step 1 — (a) Two-point formula. $m = \frac{4 - (-2)}{-1 - 3} = \frac{6}{-4} = -\frac{3}{2}$ — a falling line.

Step 2 — (b) Equal $y$'s. $m = \frac{-2 - (-2)}{7 - 3} = \frac{0}{4} = 0$ — horizontal.

Step 3 — (c) Equal $x$'s. Denominator zero — the slope is not defined: a vertical line.

Step 4 — (d) From the inclination. $m = \tan 60° = \sqrt{3}$.

Takeaway: Numerator zero → horizontal; denominator zero → vertical (no slope). Both signs matter — track them.

Example 2: Two lines at 45°

If the angle between two lines is $\frac{\pi}{4}$ and the slope of one is $\frac{1}{2}$, find the slope of the other.

Solution:

Step 1 — Set up the angle formula. With $\tan 45° = 1$: $\left|\frac{m - \frac{1}{2}}{1 + \frac{m}{2}}\right| = 1$, i.e. $\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = \pm 1$.

Step 2 — Solve the $+$ case. $m - \frac{1}{2} = 1 + \frac{m}{2}$ gives $\frac{m}{2} = \frac{3}{2}$: $m = 3$.

Step 3 — Solve the $-$ case. $m - \frac{1}{2} = -1 - \frac{m}{2}$ gives $\frac{3m}{2} = -\frac{1}{2}$: $m = -\frac{1}{3}$.

Takeaway: Two answers is geometry, not error — one line on each side of the given line makes $45°$ with it.

Example 3: Perpendicularity fixes a coordinate

The line through $(-2, 6)$ and $(4, 8)$ is perpendicular to the line through $(8, 12)$ and $(x, 24)$. Find $x$.

Solution:

Step 1 — Compute both slopes. $m_1 = \frac{8 - 6}{4 + 2} = \frac{1}{3}$; $m_2 = \frac{24 - 12}{x - 8} = \frac{12}{x - 8}$.

Step 2 — Impose $m_1 m_2 = -1$. $\frac{1}{3} \cdot \frac{12}{x - 8} = -1$, so $\frac{4}{x - 8} = -1$.

Step 3 — Solve. $x - 8 = -4$: $x = 4$.

Takeaway: Perpendicularity converts a geometric condition into a one-line equation in the unknown coordinate.

Example 4: Equidistant point on the x-axis

Find a point on the $x$-axis equidistant from $(7, 6)$ and $(3, 4)$.

Solution:

Step 1 — Name the point. On the $x$-axis it is $(x, 0)$.

Step 2 — Equate squared distances. $(x-7)^2 + 36 = (x-3)^2 + 16$.

Step 3 — Expand and solve. $x^2 - 14x + 85 = x^2 - 6x + 25$ gives $8x = 60$: $x = \frac{15}{2}$.

Step 4 — Check. $\left(\frac{15}{2}-7\right)^2 + 36 = \frac{1}{4} + 36 = \frac{145}{4}$; $\left(\frac{15}{2}-3\right)^2 + 16 = \frac{81}{4} + 16 = \frac{145}{4}$ ✓.

Takeaway: Work with squared distances — the $x^2$ terms cancel and a linear equation remains.

Example 5: Slope via a midpoint

Find the slope of the line through the origin and the midpoint of the segment joining $P(0, -4)$ and $B(8, 0)$.

Solution:

Step 1 — Midpoint. $\left(\frac{0+8}{2}, \frac{-4+0}{2}\right) = (4, -2)$.

Step 2 — Slope from the origin. $m = \frac{-2 - 0}{4 - 0} = -\frac{1}{2}$.

Takeaway: Composite constructions decompose into one formula per step — midpoint first, slope second.

Example 6: Right angle without Pythagoras

Show that $(4, 4)$, $(3, 5)$ and $(-1, -1)$ are vertices of a right-angled triangle.

Solution:

Step 1 — Slopes of two sides from $(4,4)$. To $(-1,-1)$: $\frac{-1-4}{-1-4} = 1$. To $(3,5)$: $\frac{5-4}{3-4} = -1$.

Step 2 — Multiply. $1 \times (-1) = -1$ — the two sides are perpendicular.

Step 3 — Conclude. The angle at $(4, 4)$ is a right angle. ∎

Takeaway: The slope test replaces three distance computations with two subtractions.

Example 7: Angle with the y-axis

Find the slope of the line making a $30°$ angle with the positive direction of the $y$-axis, measured anticlockwise.

Solution:

Step 1 — Convert to an inclination. The positive $y$-axis sits at $90°$ from the positive $x$-axis; anticlockwise $30°$ more gives inclination $120°$.

Step 2 — Take the tangent. $m = \tan 120° = -\tan 60° = -\sqrt{3}$.

Takeaway: Always convert to the inclination from the positive $x$-axis before applying $m = \tan\theta$.

Example 8: Parallelogram by slopes

Without the distance formula, show $(-2, -1)$, $(4, 0)$, $(3, 3)$, $(-3, 2)$ are vertices of a parallelogram.

Solution:

Step 1 — First pair of opposite sides. $(-2,-1)$ to $(4,0)$: $m = \frac{1}{6}$; $(-3,2)$ to $(3,3)$: $m = \frac{1}{6}$ — parallel.

Step 2 — Second pair. $(4,0)$ to $(3,3)$: $m = \frac{3}{-1} = -3$; $(-2,-1)$ to $(-3,2)$: $m = \frac{3}{-1} = -3$ — parallel.

Step 3 — Conclude. Two pairs of parallel opposite sides: a parallelogram. ∎

Takeaway: "Without the distance formula" is a hint to use slopes — half the arithmetic, same conclusion.

Example 9: Inclination from two points

Find the angle between the $x$-axis and the line joining $(3, -1)$ and $(4, -2)$.

Solution:

Step 1 — Slope. $m = \frac{-2 - (-1)}{4 - 3} = -1$.

Step 2 — Solve $\tan\theta = -1$ in range. With $0° \le \theta < 180°$: $\theta = 135°$.

Takeaway: Negative slope forces an obtuse inclination — never answer $-45°$; inclinations live in $[0°, 180°)$.

Example 10: Double slope

The slope of one line is double that of another. If the tangent of the angle between them is $\frac{1}{3}$, find the slopes.

Solution:

Step 1 — Set up. Slopes $m$ and $2m$: $\left|\frac{2m - m}{1 + 2m^2}\right| = \frac{1}{3}$, so $3|m| = 1 + 2m^2$.

Step 2 — Solve for $m > 0$. $2m^2 - 3m + 1 = 0$ gives $(2m - 1)(m - 1) = 0$: $m = 1$ or $\frac{1}{2}$.

Step 3 — Collect all answers. Slopes $(1, 2)$ or $\left(\frac{1}{2}, 1\right)$; the $m < 0$ sign gives the mirror pairs $(-1, -2)$ and $\left(-\frac{1}{2}, -1\right)$.

Takeaway: The modulus hides a second family of solutions — solve each sign case and report all.

Example 11: Collinearity by slope

Show that $(3, 0)$, $(-2, -2)$ and $(8, 2)$ are collinear.

Solution:

Step 1 — Two slopes from the shared point. $(3,0)$ to $(-2,-2)$: $\frac{-2}{-5} = \frac{2}{5}$; $(3,0)$ to $(8,2)$: $\frac{2}{5}$.

Step 2 — Conclude. Equal slopes through a common point put all three points on one line. ∎

Takeaway: Collinearity = equal slopes from a shared point — one comparison, no distances.