Home Class 11 Applied Maths Chapter 15: Straight Lines Exercise 15.2: Slope of a Line
Chapter 15 • Unit VII

Exercise 15.2: Slope of a Line

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

15.2 Slope of a Line: Inclination, Parallelism, Perpendicularity, and Collinearity

In Applied Mathematics, corporate financial planning, and microeconomic supply-demand curves, the slope (gradient) of a line represents the constant rate of change between two variables. Whether measuring the unit variable cost in an expense curve or determining whether two commercial flight paths are parallel, slope provides the foundational directional metric.

Fundamental Definitions & Core Theorems

  • Inclination (\(\theta\)): The angle made by a straight line with the positive direction of the \(x\)-axis, measured in the anti-clockwise direction: \[ 0^\circ \le \theta < 180^\circ \]
  • Slope (Gradient, \(m\)): \[ m = \tan \theta \quad (\theta \neq 90^\circ) \]
    • Horizontal line (parallel to \(x\)-axis): \(\theta = 0^\circ \implies m = \tan 0^\circ = 0\).
    • Vertical line (parallel to \(y\)-axis): \(\theta = 90^\circ \implies m = \tan 90^\circ\) is undefined.
    • Acute inclination (\(0^\circ < \theta < 90^\circ\)): \(m > 0\) (Upward sloping; positive economic relationship).
    • Obtuse inclination (\(90^\circ < \theta < 180^\circ\)): \(m < 0\) (Downward sloping; demand curve relationship).
  • Slope Through Two Points: The slope of the line passing through \(A(x_1, y_1)\) and \(B(x_2, y_2)\): \[ m = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_1 \neq x_2) \]
  • Condition of Parallelism: Two non-vertical lines \(L_1\) and \(L_2\) with slopes \(m_1\) and \(m_2\) are parallel if and only if their slopes are equal: \[ L_1 \parallel L_2 \iff m_1 = m_2 \]
  • Condition of Perpendicularity (Orthogonality): Two non-vertical lines \(L_1\) and \(L_2\) are perpendicular if and only if their slopes are negative reciprocals: \[ L_1 \perp L_2 \iff m_1 \cdot m_2 = -1 \iff m_2 = -\frac{1}{m_1} \]
  • Acute Angle (\(\theta\)) Between Two Intersecting Lines: \[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \quad (1 + m_1 m_2 \neq 0) \]
  • Collinearity of Three Points via Slopes: Points \(A, B, C\) lie on the same straight line if and only if: \[ \text{Slope of } AB = \text{Slope of } BC \]

Comprehensive Applied Example: Perpendicular Production Boundaries

Problem: Line \(L_1\) passes through points \(A(3, y)\) and \(B(2, 7)\), while line \(L_2\) passes through points \(C(-1, 4)\) and \(D(0, 6)\).

  1. Find the value of \(y\) if line \(L_1\) is parallel to line \(L_2\).
  2. Find the value of \(y\) if line \(L_1\) is perpendicular to line \(L_2\).
  3. Find the acute angle between \(L_2\) and the line \(y = 3x - 5\).

Solution:

First find the slope of \(L_2\):

\[ m_2 = \frac{6 - 4}{0 - (-1)} = \frac{2}{1} = 2. \]

The slope of \(L_1\) is:

\[ m_1 = \frac{7 - y}{2 - 3} = \frac{7 - y}{-1} = y - 7. \]
  1. For Parallel Lines (\(m_1 = m_2\)): \[ y - 7 = 2 \implies y = 9. \]
  2. For Perpendicular Lines (\(m_1 \cdot m_2 = -1\)): \[ (y - 7)(2) = -1 \implies 2y - 14 = -1 \implies 2y = 13 \implies y = \frac{13}{2} = 6.5. \]
  3. Acute Angle with \(y = 3x - 5\): Here \(m_1 = 3\) and \(m_2 = 2\): \[ \tan \theta = \left| \frac{3 - 2}{1 + 3(2)} \right| = \left| \frac{1}{1 + 6} \right| = \frac{1}{7} \] \[ \theta = \tan^{-1}\left(\frac{1}{7}\right) \approx 8.13^\circ. \]
Previous: Ex 15.1 Chapter Hub Next: Ex 15.3