y=mx+cCoordinate Planed=|ax₁+by₁+c|/√(a²+b²)
Chapter 9 · Class XI Mathematics · NCERT Exercises

Straight Lines — Exercises

Every Slope, Every Distance — Complete Straight Line Exercise Solutions

📂 4 Exercises 📝 70 Questions 🎓 High
▶ Start Exercises 📖 Revise Notes

Exercise Index

①
Exercise 9.1
Slope; collinearity; angle of inclination
11 Q Easy
②
Exercise 9.2
All 5 forms: slope-intercept, point-slope, two-point, intercept, normal
19 Q Moderate
③
Exercise 9.3
Parallel · perpendicular · distance; angle bisectors
17 Q Hard
★
Miscellaneous
Concurrent; locus; foot of perpendicular; family of lines
23 Q Hard

4 exercise files · 70 total questions

Chapter at a Glance

JEE MainJEE AdvancedCBSE BoardsBITSAT
14Concepts
22Formulas
HighDifficulty
8–10%Weightage

Before You Begin

Prerequisites

  • ✓Coordinate geometry — 2D (Class X)
  • ✓Distance & section formula
  • ✓Basic algebra

Have Ready

  • 🔧Graph paper
  • 🔧Ruler
  • 🔧Set square for perpendiculars

Exercise Topic Map

Exercise 9.1m=(y₂−y₁)/(x₂−x₁); collinearity; inclination θ=arctan(m)
Exercise 9.2Write in 5 forms: y=mx+c; y−y₁=m(x−x₁); two-point; x/a+y/b=1; normal
Exercise 9.3Parallel: m₁=m₂; perpendicular: m₁m₂=−1; d=|ax₁+by₁+c|/√(a²+b²)
MiscellaneousL₁+λL₂=0 family; concurrent via determinant; locus derivation

Key Formulae

\(m = \dfrac{y_2-y_1}{x_2-x_1} = \tan\theta\)
\(d = \dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\)
\(\text{Parallel: }m_1=m_2;\quad \text{Perpendicular: }m_1 m_2=-1\)
\(\tan\theta = \left|\dfrac{m_1-m_2}{1+m_1 m_2}\right|\)

NCERT Solving Method

Step 1 — Convert all lines to general form ax+by+c=0 before distance/angle formulas. Step 2 — Parallel/perpendicular: extract m from given line; apply m₁=m₂ or m₁m₂=−1. Step 3 — Family L₁+λL₂=0: use the geometric condition to find λ. Step 4 — Locus: set moving point as (h,k); derive relation; replace h→x, k→y.

Continue Your Preparation

📖 Notes Theory & concepts 🧠 MCQs Topic-wise practice 🎯 PYQs JEE · CBSE · BITSAT ⚖️ True–False Topic-wise concept check ➡️ Next Chapter Conic Sections
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