What this chapter is about
This chapter introduces the study of straight lines in a two-dimensional coordinate plane. You learn how to describe a line using algebraic equations and how different forms of these equations reveal different properties of the line. Building on coordinate geometry from earlier classes, you now explore the relationship between geometry and algebra more deeply.
The chapter develops the concept of slope (gradient) of a line and shows how slope connects to the angle a line makes with the horizontal axis. You study various standard forms of the equation of a line: slope-intercept form, point-slope form, two-point form, intercept form, and the general form. Each form is useful in different situations.
After completing this chapter, you should be able to find the equation of any line given sufficient information, determine whether lines are parallel or perpendicular, calculate the angle between two lines, and find the distance from a point to a line. These skills form the foundation for studying conic sections and three-dimensional geometry in later chapters.
Key ideas
- Slope of a line measures its steepness. If a line passes through points (x₁, y₁) and (x₂, y₂), its slope m = (y₂ − y₁)/(x₂ − x₁), provided x₁ ≠ x₂. A horizontal line has slope 0; a vertical line has undefined slope.
- Slope and angle: If a non-vertical line makes an angle θ with the positive x-axis (measured anticlockwise), then m = tan θ. This angle θ is called the inclination of the line, where 0° ≤ θ < 180°.
- Parallel lines have equal slopes. If two lines have slopes m₁ and m₂, they are parallel when m₁ = m₂.
- Perpendicular lines satisfy the condition m₁ × m₂ = −1, meaning the product of their slopes equals negative one.
- Different forms of a line's equation suit different given information: slope-intercept form when slope and y-intercept are known, point-slope form when one point and slope are known, and so on.
- General form Ax + By + C = 0 (where A and B are not both zero) can represent any straight line, including vertical lines that other forms cannot easily express.
- Distance from a point to a line has a specific formula that uses the general form of the line's equation.
Formulas and facts to remember
Slope formula: m = (y₂ − y₁)/(x₂ − x₁) Meaning: Rise divided by run between two points.
Slope-intercept form: y = mx + c Meaning: m is the slope, c is the y-intercept (where the line crosses the y-axis).
Point-slope form: y − y₁ = m(x − x₁) Meaning: The line with slope m passing through the point (x₁, y₁).
Two-point form: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁) Meaning: The line passing through two given points (x₁, y₁) and (x₂, y₂).
Intercept form: x/a + y/b = 1 Meaning: The line cutting the x-axis at (a, 0) and the y-axis at (0, b).
General form: Ax + By + C = 0 Meaning: The most general way to write any straight line.
Angle between two lines: tan θ = |m₁ − m₂|/(1 + m₁m₂), provided m₁m₂ ≠ −1. Meaning: Gives the acute angle between two non-perpendicular lines.
Distance from point (x₁, y₁) to line Ax + By + C = 0: d = |Ax₁ + By₁ + C|/√(A² + B²) Meaning: The perpendicular (shortest) distance from the point to the line.
Worked examples
Example 1: Finding the equation of a line through two points
A farmer in Punjab wants to fence a straight boundary passing through points P(2, 5) and Q(6, 13) on a coordinate map where units are in metres. Find the equation of this line.
Step 1: Calculate the slope. m = (13 − 5)/(6 − 2) = 8/4 = 2
Step 2: Use point-slope form with point P(2, 5). y − 5 = 2(x − 2) y − 5 = 2x − 4 y = 2x + 1
Answer: The equation of the boundary line is y = 2x + 1, or equivalently, 2x − y + 1 = 0.
Example 2: Checking if two roads are perpendicular
Two roads in a town are represented by lines L₁: 3x + 4y = 12 and L₂: 4x − 3y = 7. Determine whether the roads meet at right angles.
Step 1: Find slope of L₁ by rewriting in slope-intercept form. 4y = −3x + 12 y = (−3/4)x + 3 So m₁ = −3/4
Step 2: Find slope of L₂. −3y = −4x + 7 y = (4/3)x − 7/3 So m₂ = 4/3
Step 3: Check the product of slopes. m₁ × m₂ = (−3/4) × (4/3) = −1
Answer: Since the product of slopes is −1, the roads are perpendicular.
Example 3: Distance from a village to a railway line
A railway track runs along the line 5x + 12y − 60 = 0, where coordinates are in kilometres. A village is located at the point (7, 1). Find the shortest distance from the village to the track.
Step 1: Identify A = 5, B = 12, C = −60, and point (x₁, y₁) = (7, 1).
Step 2: Apply the distance formula. d = |5(7) + 12(1) + (−60)|/√(5² + 12²) d = |35 + 12 − 60|/√(25 + 144) d = |−13|/√169 d = 13/13 = 1
Answer: The shortest distance from the village to the railway track is 1 kilometre.
Common mistakes
Forgetting that vertical lines have undefined slope → Use the form x = k for vertical lines; slope formulas do not apply to them.
Using the wrong sign in the perpendicular condition, writing m₁ × m₂ = 1 → The correct condition is m₁ × m₂ = −1.
Confusing x-intercept and y-intercept in the intercept form → In x/a + y/b = 1, the line crosses the x-axis at a and the y-axis at b.
Forgetting absolute value when computing distance → The formula uses |Ax₁ + By₁ + C| because distance is always non-negative.
Mixing up which coordinate is subtracted first in the slope formula → Be consistent: m = (y₂ − y₁)/(x₂ − x₁), same order in numerator and denominator.
Quick revision
- Slope m = (y₂ − y₁)/(x₂ − x₁) = tan θ, where θ is the inclination angle.
- Parallel lines: equal slopes. Perpendicular lines: product of slopes = −1.
- Point-slope form y − y₁ = m(x − x₁) is fastest when you know one point and the slope.
- Distance from (x₁, y₁) to Ax + By + C = 0 is |Ax₁ + By₁ + C|/√(A² + B²).
- Convert any line equation to general form Ax + By + C = 0 for uniform comparison.