📘 Exercise 7.2: Slope, Inclination & Equations of Lines
Master the concepts of Slope and Inclination with detailed solutions from Chapter 7 – Coordinate Geometry. This guide provides step-by-step explanations for calculating slope, inclination, testing collinearity, determining parallel and perpendicular lines, and deriving equations of lines in various forms according to the Punjab PECTAA 2026 syllabus.
Quick Links:
Chapter 1Real Numbers
Chapter 2Logarithms
Chapter 3Sets and Functions
Chapter 4Factorization and Algebraic Manipulation
Chapter 5Linear Equations and Inequalities
Chapter 6Trigonometry
Chapter 7Coordinate Geometry
Chapter 8Logic
Chapter 9Similar Figures
Chapter 10Graphs of Functions
Chapter 11Loci and Construction
Chapter 12Information Handling
Chapter 13Probability
Exercise 7.2 - Slope, Inclination & Equations of Lines
This exercise focuses on slope, inclination, collinearity conditions, parallel/perpendicular lines, and various forms of line equations including slope-intercept, two-intercept, and normal form.
Exercise 7.2 builds essential skills for analytical geometry: calculating slope between two points, finding inclination angle, testing collinearity using equal slopes, determining if lines are parallel or perpendicular, and deriving line equations from given conditions. Students also learn to convert line equations between slope-intercept form, two-intercept form, and normal form.
Key Topics Covered
- ✅ Slope Formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
- ✅ Inclination: \(\tan \alpha = m\)
- ✅ Collinearity using equal slopes
- ✅ Parallel lines condition: \(m_1 = m_2\)
- ✅ Perpendicular lines condition: \(m_1 m_2 = -1\)
- ✅ Equation of line: point-slope form, slope-intercept form (\(y = mx + c\)), two-intercept form (\(\frac{x}{a} + \frac{y}{b} = 1\)), normal form (\(x\cos\alpha + y\sin\alpha = p\))
- ✅ Perpendicular bisector of a segment
📐 Important Note:
Remember that slope is undefined for vertical lines (infinite slope). For collinear points, slopes between any two pairs are equal. For perpendicular lines, product of slopes equals -1 (provided both slopes are defined).
📌 Sample Problem: Slope & Inclination
Q1 (i): Find slope and inclination of line joining (-2,4) and (5,11).
\(m = \frac{11-4}{5-(-2)} = \frac{7}{7} = 1\) → \(\alpha = \tan^{-1}(1) = 45^\circ\)
Q3: Find \(k\) so that line joining A(7,3), B(k,-6) and line joining C(-4,5), D(-6,4) are parallel/perpendicular.
Slope of AB = \(\frac{-9}{k-7}\), slope of CD = \(\frac{1}{2}\). For parallel: \(\frac{-9}{k-7} = \frac{1}{2} \Rightarrow k = -11\). For perpendicular: \(\left(\frac{-9}{k-7}\right)\left(\frac{1}{2}\right) = -1 \Rightarrow k = \frac{23}{2}\).
Q10 (a): Convert \(2x - 4y + 11 = 0\) to slope-intercept, two-intercept, and normal form.
Slope-intercept: \(y = \frac{x}{2} + \frac{11}{4}\); Two-intercept: \(\frac{x}{-11/2} + \frac{y}{11/4} = 1\); Normal form: \(x\cos116.56^\circ + y\sin116.56^\circ = \frac{11}{2\sqrt{5}}\).
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus