📘 Exercise 7.1: Basics of Plane Analytical Geometry
Master the foundational concepts of Distance and Ratio Formulas with detailed solutions from Chapter 7 – Plane Analytical Geometry. This guide provides step-by-step explanations for calculating the distance between points, finding midpoints, and applying the Ratio Formula for internal and external division 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 7.1 - Coordinate Geometry (Plane Analytical Geometry)
This exercise introduces fundamental concepts of plane analytical geometry: distance between two points, midpoint of a segment, collinearity conditions, and geometric properties of shapes using coordinate methods.
Exercise 7.1 builds a strong foundation for coordinate geometry by applying the distance formula and midpoint formula to solve real geometric problems. Students learn to identify collinear points, verify right triangles, isosceles triangles, and parallelograms using algebraic methods.
Key Topics Covered
- ✅ Distance Formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
- ✅ Midpoint Formula: \(M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)\)
- ✅ Collinear points condition using equal slopes or distance relations
- ✅ Verifying right triangles via Pythagorean theorem
- ✅ Proving isosceles triangles (two equal sides)
- ✅ Parallelogram verification via midpoint of diagonals
- ✅ Finding centre and radius of a circle given endpoints of diameter
📐 Important Note:
Remember that the distance formula is derived from the Pythagorean theorem. The midpoint formula gives the point exactly halfway between two endpoints. For collinear points, the slope between any two points remains constant.
📌 Sample Problem: Distance between Points
Q2 (i): Find distance between A(6,7) and B(0,-2).
\(|AB| = \sqrt{(0-6)^2 + (-2-7)^2} = \sqrt{(-6)^2+(-9)^2} = \sqrt{36+81}=\sqrt{117}=3\sqrt{13}\).
Midpoint (Q3a): A(3,1), B(-2,-4): \(M = \left(\frac{3+(-2)}{2},\frac{1+(-4)}{2}\right)=\left(\frac12,-\frac32\right)\).
Collinearity (Q7): Find \(h\) if A(-1,h), B(3,2), C(7,3) are collinear → slope AB = slope BC → \(\frac{2-h}{4}=\frac14 \Rightarrow h=1\).
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus