Equation of a Straight Line — chapter notes

• The inclination $\theta$ of a line is the angle from the positive $x$-axis to the line, measured anticlockwise, with $0^\circ \le \theta < 180^\circ$.
• The slope (gradient) is $m = \tan \theta$.
• $m > 0$: the line rises from left to right; $m < 0$: it falls (obtuse $\theta$); $m = 0$: horizontal; a vertical line ($\theta = 90^\circ$) has no slope.

Example: a line inclined at $60^\circ$ has slope $\sqrt{3}$; a line of slope $-1$ has inclination $135^\circ$.

• For $(x_1, y_1)$ and $(x_2, y_2)$ with $x_1 \ne x_2$: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$, the rise divided by the run. Keep the same order on top and bottom.
• If $x_1 = x_2$ the line is vertical and has no slope.
• $A$, $B$, $C$ are collinear when slope $AB$ = slope $BC$.

Example: $P(-2, 5)$ and $Q(4, -1)$: $m = \dfrac{-1 - 5}{4 - (-2)} = -1$. For $A(1, -1)$, $B(3, 3)$, $C(k, 7)$ to be collinear, $\dfrac{7 - 3}{k - 3} = 2$, so $k = 5$.

• $y = mx + c$: slope $m$, $y$-intercept $c$ (the line meets the $y$-axis at $(0, c)$). A line through the origin is $y = mx$.
• From $ax + by + c = 0$ ($b \ne 0$): $y = -\dfrac{a}{b}x - \dfrac{c}{b}$, so the slope is $-\dfrac{a}{b}$ and the $y$-intercept is $-\dfrac{c}{b}$.
• Intercepts: put $y = 0$ for the $x$-intercept and $x = 0$ for the $y$-intercept.
• A point lies on a line exactly when its coordinates satisfy the equation.

Example: $3x - 4y + 8 = 0$ gives $y = \dfrac{3}{4}x + 2$: slope $\dfrac{3}{4}$, $y$-intercept $2$. The line $4x + 3y = 12$ meets the axes at $(3, 0)$ and $(0, 4)$.

• Point-slope form: the line through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
• Two-point form: find $m = \dfrac{y_2 - y_1}{x_2 - x_1}$, then use the point-slope form with either point.
• Clear fractions and write the answer as $ax + by + c = 0$ (or $y = mx + c$); check it with the given points.
• Special cases: a line through $(x_1, y_1)$ parallel to the $x$-axis is $y = y_1$; parallel to the $y$-axis, $x = x_1$.

Example: through $(2, -3)$ with slope $\dfrac{3}{2}$: $2y + 6 = 3x - 6$, so $3x - 2y - 12 = 0$. Through $(-1, 4)$ and $(3, -2)$: $m = -\dfrac{3}{2}$ and $3x + 2y - 5 = 0$.

• Parallel: $m_1 = m_2$ (and different intercepts). Two vertical lines are parallel.
• Perpendicular: $m_1 m_2 = -1$, so the perpendicular slope is $-\dfrac{1}{m}$. A horizontal and a vertical line are perpendicular.
• A line parallel to $ax + by + c = 0$ can be written $ax + by + k = 0$; find $k$ from the point it passes through.
• Unknown coefficients: set the slopes equal (parallel) or their product equal to $-1$ (perpendicular).

Example: through $(3, -2)$ parallel to $4x - y + 7 = 0$: $4x - y - 14 = 0$. Through $(-2, 1)$ perpendicular to $x + 3y - 6 = 0$: slope $3$, so $3x - y + 7 = 0$.

• The point where two lines meet satisfies both equations: solve them simultaneously by substitution or elimination.
• To find a line through the meeting point with a given condition, find the point first, then the slope, then use the point-slope form.
• The foot of the perpendicular from a point to a line is where the perpendicular through the point meets the line.

Example: $2x + y - 7 = 0$ and $x - y + 1 = 0$ meet at $(2, 3)$; the line through $(2, 3)$ perpendicular to $3x - 4y + 2 = 0$ is $4x + 3y - 17 = 0$.

• Perpendicular bisector of $AB$: through the mid-point of $AB$, slope $-\dfrac{1}{m_{AB}}$. Every point on it is equidistant from $A$ and $B$.
• Median from $A$: the line through $A$ and the mid-point of $BC$. The three medians meet at the centroid.
• Altitude from $A$: through $A$, perpendicular to $BC$ (slope $-\dfrac{1}{m_{BC}}$).
• With the section formula, a point dividing $AB$ in $m : n$ is $\left(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\right)$.

Example: for $A(-1, 3)$ and $B(5, -1)$ the perpendicular bisector is $3x - 2y - 4 = 0$; for $P(-2, -1)$, $Q(4, 1)$, $R(2, 5)$ the median from $R$ is $5x - y - 5 = 0$ and the altitude from $P$ is $x - 2y = 0$.