Class 11 — Mathematics (NCERT)

Straight Lines

Class 11

  • ✓Define the slope (gradient) of a line and calculate it using various methods.
  • ✓Determine the angle between two lines.
  • ✓Derive and apply different forms of the equation of a straight line (point-slope, two-point, slope-intercept, intercept, normal).
  • ✓Convert between different forms of the equation of a line.
  • ✓Understand the general equation of a line and its relation to other forms.

Key concepts

Slope of a Line

The slope (or gradient) of a non-vertical line is a measure of its steepness. It is denoted by 'm'.\n\n1. **Angle of Inclination (θ)**: The angle θ (0 ≤ θ < π) made by a line with the positive direction of the x-axis, measured anti-clockwise, is called its inclination.\n2. **Slope (m)**: If θ is the inclination of a line, then its slope m is given by m = tan θ. If θ = π/2 (i.e., the line is vertical), the slope is undefined.\n3. **Slope of a line passing through two points (x₁, y₁) and (x₂, y₂)**: If x₁ ≠ x₂, the slope m is given by the ratio of the difference in y-coordinates to the difference in x-coordinates.\n4. **Condition for Parallel Lines**: Two non-vertical lines are parallel if and only if their slopes are equal (m₁ = m₂).\n5. **Condition for Perpendicular Lines**: Two non-vertical lines are perpendicular if and only if the product of their slopes is -1 (m₁m₂ = -1). If one line is vertical, the other must be horizontal.\n6. **Angle between two lines**: If θ is the acute angle between two lines with slopes m₁ and m₂, then tan θ = |(m₂ - m₁) / (1 + m₁m₂)|, provided 1 + m₁m₂ ≠ 0.

m = tan θ OR m = (y₂ - y₁) / (x₂ - x₁)
Forms of the Equation of a Line

The equation of a straight line can be expressed in various forms depending on the given information.\n\n1. **Horizontal and Vertical Lines**:\n * Equation of a horizontal line (parallel to x-axis) at a distance 'k' from the x-axis is y = k.\n * Equation of a vertical line (parallel to y-axis) at a distance 'k' from the y-axis is x = k.\n2. **Point-Slope Form**: The equation of a line passing through a point (x₁, y₁) with slope 'm' is given by:\n3. **Two-Point Form**: The equation of a line passing through two distinct points (x₁, y₁) and (x₂, y₂) is given by:\n4. **Slope-Intercept Form**: The equation of a line with slope 'm' and y-intercept 'c' is given by:\n5. **Intercept Form**: The equation of a line making x-intercept 'a' and y-intercept 'b' is given by:\n6. **Normal Form**: The equation of a line where 'p' is the length of the perpendicular from the origin to the line, and 'α' is the angle which the normal (perpendicular) makes with the positive direction of the x-axis, is given by:\n7. **General Equation of a Line**: Any equation of the form Ax + By + C = 0, where A, B, C are real numbers and A and B are not both zero, represents a straight line. From this form:\n * Slope m = -A/B (if B ≠ 0)\n * x-intercept = -C/A (if A ≠ 0)\n * y-intercept = -C/B (if B ≠ 0)

1. y = k (horizontal), x = k (vertical)\n2. y - y₁ = m(x - x₁)\n3. (y - y₁) / (y₂ - y₁) = (x - x₁) / (x₂ - x₁)\n4. y = mx + c\n5. x/a + y/b = 1\n6. x cos α + y sin α = p\n7. Ax + By + C = 0

Key facts to remember

  • 1Slope of a line m = tan θ, where θ is the angle of inclination.
  • 2Slope of a line through (x₁, y₁) and (x₂, y₂) is m = (y₂ - y₁) / (x₂ - x₁).
  • 3Parallel lines have equal slopes (m₁ = m₂).
  • 4Perpendicular lines have slopes whose product is -1 (m₁m₂ = -1).
  • 5Point-slope form: y - y₁ = m(x - x₁).
  • 6Slope-intercept form: y = mx + c.
  • 7Intercept form: x/a + y/b = 1.
  • 8Normal form: x cos α + y sin α = p.
  • 9General equation of a line: Ax + By + C = 0.

Worked examples

Example 1

Find the slope of the line passing through the points (3, -2) and (-1, 4). Also, find the equation of this line.

ILet (x₁, y₁) = (3, -2) and (x₂, y₂) = (-1, 4).
IIThe slope 'm' is given by m = (y₂ - y₁) / (x₂ - x₁).
IIISubstitute the coordinates: m = (4 - (-2)) / (-1 - 3) = (4 + 2) / (-4) = 6 / (-4) = -3/2.
IVNow, to find the equation of the line, we can use the point-slope form: y - y₁ = m(x - x₁).
VUsing point (3, -2) and m = -3/2:
VIy - (-2) = (-3/2)(x - 3)
VIIy + 2 = (-3/2)(x - 3)
VIIIMultiply both sides by 2: 2(y + 2) = -3(x - 3)
92y + 4 = -3x + 9
10Rearrange to the general form: 3x + 2y + 4 - 9 = 0
113x + 2y - 5 = 0

Answer

The slope of the line is -3/2. The equation of the line is 3x + 2y - 5 = 0.

You can verify the equation by substituting the other point (-1, 4). 3(-1) + 2(4) - 5 = -3 + 8 - 5 = 0.

Example 2

Find the equation of the line which makes intercepts -3 and 2 on the x and y axes respectively. Also, find the angle it makes with the positive x-axis.

IGiven x-intercept a = -3 and y-intercept b = 2.
IIUsing the intercept form of the equation of a line: x/a + y/b = 1.
IIISubstitute the values: x/(-3) + y/2 = 1.
IVTo remove denominators, find the L.C.M. of 3 and 2, which is 6. Multiply the entire equation by 6:
V6(x/(-3)) + 6(y/2) = 6(1)
VI-2x + 3y = 6
VIIRearrange to general form: 2x - 3y + 6 = 0.
VIIITo find the angle with the x-axis, we need the slope. Convert the equation to slope-intercept form (y = mx + c):
93y = 2x + 6
10y = (2/3)x + 2.
11Comparing with y = mx + c, the slope m = 2/3.
12The angle θ made with the positive x-axis is given by tan θ = m.
13tan θ = 2/3.
14θ = tan⁻¹(2/3).

Answer

The equation of the line is 2x - 3y + 6 = 0. The angle it makes with the positive x-axis is tan⁻¹(2/3).

The angle tan⁻¹(2/3) is an acute angle, as the slope is positive.

Example 3

Reduce the equation 3x + 4y - 12 = 0 to normal form and find the perpendicular distance from the origin and the angle of the normal.

IThe given equation is 3x + 4y - 12 = 0.
IIFirst, move the constant term to the R.H.S. and ensure it is positive: 3x + 4y = 12.
IIICompare this with Ax + By = -C. Here A = 3, B = 4, C = -12.
IVCalculate √(A² + B²) = √(3² + 4²) = √(9 + 16) = √25 = 5.
VDivide the entire equation by √(A² + B²) = 5:
VI(3/5)x + (4/5)y = 12/5.
VIIThis is the normal form x cos α + y sin α = p.
VIIIComparing, we get: p = 12/5, cos α = 3/5, sin α = 4/5.
9Since both cos α and sin α are positive, α lies in the first quadrant.
10α = cos⁻¹(3/5) or α = sin⁻¹(4/5). (This value can be left in inverse trigonometric form or approximated if a calculator is allowed).
11The perpendicular distance from the origin is p = 12/5 units.

Answer

The normal form of the equation is (3/5)x + (4/5)y = 12/5. The perpendicular distance from the origin is 12/5 units, and the angle of the normal with the positive x-axis is α = cos⁻¹(3/5).

Always ensure the constant term 'p' in the normal form is positive. If it's negative after moving to RHS, divide by -√(A²+B²).

Common mistakes

  • ✗Confusing the order of coordinates in the slope formula, e.g., (y₁ - y₂) / (x₂ - x₁).
  • ✗Incorrectly identifying 'a' as y-intercept and 'b' as x-intercept in the intercept form.
  • ✗Forgetting to ensure the constant term 'p' is positive in the normal form equation.
  • ✗Errors in algebraic manipulation when converting between different forms of the line equation.
  • ✗Not considering the special cases of vertical lines (undefined slope) and horizontal lines (zero slope).

Exam tips

  • ★Always draw a rough sketch of the line and given points if possible; it helps in visualising the problem and checking your answer.
  • ★Memorise all the different forms of the equation of a straight line and their specific uses.
  • ★Pay close attention to signs, especially when substituting coordinates or rearranging equations.
  • ★Practise converting an equation from one form to another, as this is a common type of question.
  • ★Show all steps clearly and logically in your solutions to avoid losing marks for calculation errors.

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