Home Class 11 Applied Maths Chapter 15: Straight Lines Exercise 15.4: Distance Formulas
Chapter 15 • Unit VII

Exercise 15.4: Distance Formulas

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

15.4 Distance Formulas: Point-to-Line and Parallel Line Clearances

In Applied Mathematics, civil engineering, regional urban planning, and spatial logistics, calculating the shortest distance between facilities and linear infrastructures (highways, power lines, transit tracks) is a critical optimization task. The perpendicular distance formulas translate spatial separation into direct algebraic computation.

Fundamental Formulae & Theorems

  • Perpendicular Distance of a Point from a Line: The perpendicular distance \(d\) from a point \(P(x_1, y_1)\) to the line \(Ax + By + C = 0\): \[ d = \frac{|A x_1 + B y_1 + C|}{\sqrt{A^2 + B^2}} \]
  • Distance from the Origin \((0, 0)\): \[ d_0 = \frac{|C|}{\sqrt{A^2 + B^2}} \]
  • Distance Between Two Parallel Lines: Two parallel lines can always be written with identical \(x\) and \(y\) coefficients: \[ L_1: Ax + By + C_1 = 0 \quad \text{and} \quad L_2: Ax + By + C_2 = 0 \] The perpendicular distance \(d\) between \(L_1\) and \(L_2\) is: \[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \] Slope-Intercept Form: If \(y = mx + c_1\) and \(y = mx + c_2\): \[ d = \frac{|c_1 - c_2|}{\sqrt{1 + m^2}} \]

Comprehensive Applied Example: Hazardous Pipeline Buffer Clearance

Scenario: A high-pressure natural gas pipeline runs along the straight line \(3x - 4y + 10 = 0\) across an industrial zone. A petrochemical warehouse is located at coordinates \(P(4, -1)\).

  1. Find the shortest distance between the warehouse and the pipeline.
  2. If municipal safety regulations mandate a minimum safety clearance buffer of 6 km, determine whether the warehouse meets regulatory compliance.
  3. A parallel reserve pipeline is planned along the line \(6x - 8y - 15 = 0\). What is the exact separation distance between the primary and reserve pipelines?

Solution:

  1. Distance from Warehouse \(P(4, -1)\) to Pipeline: Here \(A = 3\), \(B = -4\), \(C = 10\), \(x_1 = 4\), \(y_1 = -1\): \[ d = \frac{|3(4) + (-4)(-1) + 10|}{\sqrt{3^2 + (-4)^2}} = \frac{|12 + 4 + 10|}{\sqrt{9 + 16}} = \frac{|26|}{\sqrt{25}} = \frac{26}{5} = 5.2 \text{ km.} \]
  2. Compliance Verification: Since the actual distance is \(5.2\text{ km} < 6\text{ km}\), the warehouse violates the 6 km safety clearance buffer.
  3. Separation Distance Between Parallel Pipelines: The primary line is \(3x - 4y + 10 = 0\). Multiply by 2 so coefficients match: \[ 6x - 8y + 20 = 0 \quad \implies C_1 = 20 \] The secondary line is: \[ 6x - 8y - 15 = 0 \quad \implies C_2 = -15 \] Now apply the parallel line distance formula: \[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} = \frac{|20 - (-15)|}{\sqrt{6^2 + (-8)^2}} = \frac{|35|}{\sqrt{36 + 64}} = \frac{35}{\sqrt{100}} = \frac{35}{10} = 3.5 \text{ km.} \] Separation Distance = 3.5 km.
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