Home Class 11 Applied Maths Chapter 15: Straight Lines Exercise 15.3: Various Forms of Equations of a Line
Chapter 15 • Unit VII

Exercise 15.3: Various Forms of Equations of a Line

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

15.3 Various Forms of Equations of a Line: Point-Slope, Intercept, and Business Models

In Applied Mathematics, corporate budgeting, and econometric regression, straight lines are the primary mathematical instrument for modeling linear trends. From linear cost accounting ($C = mx + F$) to asset depreciation schedules and consumer demand relationships ($p = -mx + c$), mastering the various forms of straight line equations allows analysts to translate empirical data into predictive algebraic models.

Master Formulary of Straight Line Equations

Standard Form Given Geometric Parameters Algebraic Equation
Horizontal Line Parallel to \(x\)-axis at distance \(b\) \[ y = b \quad (\text{or } y = -b) \]
Vertical Line Parallel to \(y\)-axis at distance \(a\) \[ x = a \quad (\text{or } x = -a) \]
Point-Slope Form Passes through \((x_0, y_0)\) with slope \(m\) \[ y - y_0 = m(x - x_0) \]
Two-Point Form Passes through \((x_1, y_1)\) and \((x_2, y_2)\) \[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \]
Slope-Intercept Form Slope \(m\) and \(y\)-intercept \(c\) \[ y = mx + c \]
Intercept Form \(x\)-intercept \(a\) and \(y\)-intercept \(b\) \[ \frac{x}{a} + \frac{y}{b} = 1 \]
General Form Arbitrary linear coefficients \(A, B, C\) \[ Ax + By + C = 0 \implies m = -\frac{A}{B}, \, c = -\frac{C}{B} \]

Comprehensive Applied Business Example: Linear Cost Accounting

Scenario: A manufacturing firm incurs total production costs of ₹2,50,000 when manufacturing 500 units of a product, and ₹3,70,000 when manufacturing 800 units. Assuming a linear relationship exists between output \(x\) and total cost \(C\):

  1. Find the linear cost equation using the Two-Point Form.
  2. Identify the fixed overhead cost \(F\) and the marginal variable cost per unit \(m\).
  3. Forecast the total cost of producing 1,200 units.
  4. If the firm sells each unit for ₹500, determine the break-even production volume.

Solution:

  1. Two-Point Form of Linear Cost Equation: Data points: \((x_1, C_1) = (500, 250000)\) and \((x_2, C_2) = (800, 370000)\). \[ m = \frac{C_2 - C_1}{x_2 - x_1} = \frac{3{,}70{,}000 - 2{,}50{,}000}{800 - 500} = \frac{1{,}20{,}000}{300} = 400. \] Using Point-Slope Form: \[ C - 2{,}50{,}000 = 400(x - 500) \] \[ C - 2{,}50{,}000 = 400x - 2{,}00{,}000 \implies C(x) = 400x + 50{,}000. \]
  2. Economic Metrics:
    • Marginal Variable Cost per unit: \(m = ₹400\) (the slope).
    • Fixed Overhead Cost: \(F = ₹50{,}000\) (the vertical intercept).
  3. Cost for 1,200 units: \[ C(1200) = 400(1200) + 50{,}000 = 4{,}80{,}000 + 50{,}000 = ₹5{,}30{,}000. \]
  4. Break-Even Production Volume: Total Revenue \(R(x) = 500x\). At break-even, \(R(x) = C(x)\): \[ 500x = 400x + 50{,}000 \implies 100x = 50{,}000 \implies x = 500 \text{ units.} \] Break-Even Volume = 500 units.
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