Slope and Equations of Straight Lines: Quick Revision

A brief note on Slope and Equations of Straight Lines. Read it before a test. Try each question at the end before you open its answer.

Slope and angle

• The slope through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \dfrac{y_2 - y_1}{x_2 - x_1}$.
• $m = \tan\theta$, where $\theta$ is the angle of inclination, from $0^\circ$ up to $180^\circ$.
• Positive slope: acute angle. Negative slope: obtuse angle. A vertical line has no slope.
• The acute angle between two lines has $\tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1 m_2}\right|$.

Parallel, perpendicular and collinear

• Parallel lines: $m_1 = m_2$. Perpendicular lines: $m_1 m_2 = -1$.
• A horizontal line and a vertical line are perpendicular.
• $A$, $B$, $C$ are collinear when the slope of $AB$ equals the slope of $BC$.

Forms of the equation

• Horizontal: $y = b$. Vertical: $x = a$.
• Point-slope: $y - y_1 = m(x - x_1)$. Slope-intercept: $y = mx + c$.
• Two-point: $y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
• Intercept form: $\dfrac{x}{a} + \dfrac{y}{b} = 1$, with slope $-\dfrac{b}{a}$.
• The point dividing $AB$ in the ratio $m : n$ is $\left(\dfrac{m x_2 + n x_1}{m + n}, \dfrac{m y_2 + n y_1}{m + n}\right)$.

A worked example

Find the equation of the line through $A(2, 3)$ and $B(6, 11)$, and write it in intercept form.

1. The slope. $m = \dfrac{11 - 3}{6 - 2} = 2$.
2. Point-slope form. $y - 3 = 2(x - 2)$, so $y = 2x - 1$.
3. The intercepts. At $x = 0$, $y = -1$. At $y = 0$, $x = \dfrac{1}{2}$.
4. Intercept form. $\dfrac{x}{\frac{1}{2}} + \dfrac{y}{-1} = 1$.

Practice with new numbers

A program to try

The program finds the slope, the angle, the intercepts and the intercept form of the line through two points, and checks the form with one of the points.

Mistakes to avoid

• Run over rise. The slope has the rise on top.
• A vertical line with slope $0$. A vertical line has no slope, and a horizontal line has slope $0$.
• Opposite slopes for perpendicular lines. The product must be $-1$.
• A plus sign in point-slope form. It is $y - y_1 = m(x - x_1)$.
• Intercept form for a line through the origin. It needs two non-zero intercepts.

Check yourself

1. Find the slope of the line through $(5, -1)$ and $(2, 8)$.

2. Find the acute angle between lines of slopes $0$ and $\sqrt{3}$.

3. Find $k$ so that $(1, 2)$, $(3, k)$ and $(5, 10)$ are collinear.

4. Write the equation of the line with slope $-2$ through $(1, 5)$.

5. A line cuts the axes at $(3, 0)$ and $(0, 6)$. Find its slope.