Line Equation and Slope

With the two-point formula $m =\frac{ (y_{2} - y_{1})}{(x_{2} - x_{1})}$: y-difference over x-difference, in the same point order. Example: P(1|2) and Q(4|11) give $m =\frac{ (11 - 2)}{(4 - 1)} =\frac{ 9}{3} = 3$; the line climbs 3 units per step to the right. Mind two traps: first, Δy belongs in the numerator (not Δx), second, you must not mix the order, otherwise the sign flips. Intuitively m is the slope triangle: 1 to the right, m upwards. For $x_{1} = x_{2}$ the formula is undefined; that is the vertical line, which has no slope.

The line falls: per step to the right it drops by |m| units. So $y = -2x + 7$ loses 2 units per x-step. In applied contexts a negative slope describes decrease processes, e.g. a water level sinking 2 cm per hour, or a battery losing 0.5 % per minute; m then carries the unit y-unit per x-unit. Fully sorted: $m > 0$ rising, $m = 0$ horizontal (constant function), $m < 0$ falling; the larger |m|, the steeper. The angle of inclination is measured negative for negative m accordingly: $\tan{} \theta{} = m$, for $m = -1$ that is −45°.

Three routes. First, read it off: b is the y-value where the line crosses the y-axis, i.e. f(0). Second, compute: if m is known and a point (x₁|y₁) is given, b follows from substituting into $y = mx + b$, rearranged $b = y_{1} - m\cdot{}x_{1}$. Example: $m = 3$ and P(1|2) give $b = 2 - 3\cdot{}1 = -1$, so $y = 3x - 1$. Third, work without b: the point-slope form $y = m(x - x_{1}) + y_{1}$ yields the line directly. Finally check with the second point if available. Do not confuse b with the root: b lies on the y-axis, the root $x_{0} = -\frac{b}{m}$ on the x-axis.

Lines are parallel exactly when their slopes agree: m₁ = m₂ (and b differs, otherwise they are identical). They are perpendicular when $m_{1}\cdot{}m_{2} = -1$; the second slope is the negative reciprocal of the first. Example: to $y = 2x + 1$ every line with $m = 2$ is parallel and every line with $m = -\frac{1}{2}$ perpendicular, e.g. $y = -0.5x + 4$. This rule is the tool for normals in calculus: the normal at a curve point has slope $-\frac{1}{f}$′(x₀). Special case: horizontal $(m = 0)$ and vertical lines are also perpendicular to each other, although the product rule formally fails there because the vertical line has no slope.

The angle of inclination θ between line and x-axis satisfies $\tan{} \theta{} = m$, so $\theta{} = \arctan{}(m)$. For $m = 1$ that is 45°, for $m = 3$ about 71.6°. Traffic signs state gradients in percent: 12 % means $m = 0.12$, i.e. 12 m of height gain per 100 m horizontal distance; the angle is $\arctan{}(0.12) \approx{} 6.8$°. Conversely 100 % corresponds to exactly 45°, not 90°; that is a popular trick question. Note that percentage and angle are not proportional, because the tangent does not grow linearly. In applied tasks it pays to distinguish consistently between m (ratio), angle (degrees) and percent $(m\cdot{}100)$.