Lines and Angles - NCERT Solutions (Class 9, Chapter 6)

Introduction

This chapter studies properties of angles formed when lines intersect and when a line intersects parallel lines. These properties are used to prove statements using deductive reasoning. You will encounter linear pairs, vertically opposite angles, corresponding and alternate angles, and properties of parallel lines.

Why it matters: Architects, physicists (ray diagrams), and engineers use these angle properties to construct models, draw plans and analyse forces and light rays.

Basic Terms & Definitions

Line-segment: Part of a line with two end points.

Ray: Part of a line with one endpoint.

Collinear points: Points that lie on the same line.

Angle: Formed by two rays with a common endpoint (vertex). Arms are the rays.

Acute angle: 0° < θ < 90°
Right angle: θ = 90°
Obtuse angle: 90° < θ < 180°
Straight angle: θ = 180°
Reflex angle: 180° < θ < 360°

Adjacent angles: Share a common vertex and common arm; non-common arms are on different sides of common arm.

Linear pair: Two adjacent angles whose non-common arms form a line; they sum to 180°.

Vertically opposite angles: When two lines intersect, the opposite angles are equal.

Axioms & Theorems

Linear Pair Axiom

Axiom 6.1: If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.

Axiom 6.2 (Converse): If the sum of two adjacent angles is 180°, then the non-common arms form a line.

Theorem (Vertically Opposite Angles)

If two lines intersect, the vertically opposite angles are equal. (Proof uses linear pair axiom.)

Parallel lines

If two lines are parallel to the same line, they are parallel to each other.

Important Examples

Example 1: In a figure where lines PQ and RS intersect at O, if ∠POR : ∠ROQ = 5 : 7, find all angles at O.

Since ∠POR + ∠ROQ = 180° and ratio is 5:7, total parts = 12.

∠POR = (5/12) × 180° = 75°; ∠ROQ = (7/12) × 180° = 105°.

Vertically opposite angles give the remaining two: ∠POS = 105°, ∠SOQ = 75°.

Example 2: Ray OS stands on a line POQ. OR bisects ∠POS and OT bisects ∠SOQ. If ∠POS = x, find ∠ROT.

∠SOQ = 180° − x. OR bisects ∠POS ⇒ ∠ROS = x/2. OT bisects ∠SOQ ⇒ ∠SOT = (180° − x)/2 = 90° − x/2.

Thus ∠ROT = ∠ROS + ∠SOT = x/2 + (90° − x/2) = 90°.

Example 3: OP, OQ, OR and OS are four rays. Prove ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°.

Produce OQ backwards to T so TOQ is a line. Then OP stands on TOQ ⇒ ∠TOP + ∠POQ = 180°. Similarly, OS stands on TOQ ⇒ ∠TOS + ∠SOQ = 180°.

But ∠SOQ = ∠SOR + ∠QOR. Add the two equations and simplify to get the sum = 360°.

Example 4: If PQ ∥ RS, ∠MXQ = 135° and ∠MYR = 40°, find ∠XMY.

Draw AB through M parallel to PQ (so AB ∥ PQ ∥ RS). From interior angle relations, ∠XMB = 45° and ∠BMY = 40°. So ∠XMY = 45° + 40° = 85°.

Example 5: If a transversal intersects two lines such that bisectors of a pair of corresponding angles are parallel, prove the two lines are parallel.

Let BE and CG be bisectors and BE ∥ CG. Then ∠ABE/2 = ∠BCG/2 as corresponding angles, so ∠ABQ = ∠BCS; hence corresponding angles are equal ⇒ lines are parallel.

Example 6: AB ∥ CD ∥ EF and EA ⟂ AB. If ∠BEF = 55°, find x, y and z (angles shown in diagram).

y + 55° = 180° ⇒ y = 125°. x = y (corresponding) ⇒ x = 125°. For z, 90° + z + 55° = 180° ⇒ z = 35°.

Exercise 6.1 — Questions & Answers

1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution: Let us compute step by step. From intersection, vertically opposite and linear pair relations apply. (Answer depends on diagram labels; typical solution process: express angles using linear pairs and equalities, solve for ∠BOE, then reflex ∠COE = 360° − ∠COE (or compute from found acute/obtuse angle)).

(Note: For the printable page include the figure; here we give the method: use ∠AOC + ∠COB = 180°, relate given sums and substitute.)

2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Solution: Use linear pair and proportional parts. If a:b = 2:3, total 5 parts correspond to 180° − 90° = 90° (depending on orientation). Then c can be computed from the parts — explain and calculate based on figure.

(Diagram-specific; apply interior/exterior angle relations and solve.)

3. In Fig. 6.15, if ∠PQR = ∠PRQ, prove that ∠PQS = ∠PRT.

Since ∠PQR = ∠PRQ, triangle PQR is isosceles (PQ = PR). Using properties of parallel lines or equal alternate angles (depending on S and T placement), show corresponding equal angles. Conclude ∠PQS = ∠PRT.

4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Linear pair axiom: If sum of adjacent angles is 180°, then non-common arms form a line. Here x + y = w + z implies the adjacent sums equal 180°, hence A, O, B are collinear.

5. In Fig. 6.17, POQ is a line. Ray OR ⟂ PQ. OS lies between OP and OR. Prove that ∠ROS = ½(∠QOS − ∠POS).

Sketch: Use that OR is perpendicular to PQ, and write ∠QOS = ∠QOR + ∠ROS and ∠POS = ∠POR + ∠ROS. Subtract and rearrange to isolate ∠ROS. Using bisector relations or dividing by 2 yields the result.

6. It is given that ∠XYZ = 64° and XY is produced to P. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Since XY produced to P gives ∠ZYP = 180° − 64° = 116°. Ray YQ bisects it so each half = 58°. Thus ∠XYQ = 58° − (angle between XY and YZ) depending on orientation; reflex ∠QYP = 360° − 116° = 244° (or compute from constructed parts).

Notes: The NCERT exercise 6.1 questions rely on accompanying diagrams (Fig. 6.13–6.17). The solution steps above show method and key computations; for each specific diagram, substitute the labelled angles to compute numeric values.

Exercise 6.2 — Questions & Answers

1. In Fig. 6.23, if AB ∥ CD, CD ∥ EF and y : z = 3 : 7, find x.

Use parallel-line relations: corresponding and interior angles. If y:z = 3:7 and lines are parallel, express x in terms of parts of 180° (or 360°) as indicated by diagram. Solve proportionally to find x.

2. In Fig. 6.24, if AB ∥ CD, EF ⟂ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

EF ⟂ CD ⇒ EF ⟂ AB. Use perpendicular and parallel relations: If ∠GED = 126°, compute supplementary and complementary angles to get required values. For example ∠AGE = 54° (if it is supplementary), ∠GEF = 90° − (related angle), etc. (Follow figure-specific relations to compute exact numbers.)

3. In Fig. 6.25, if PQ ∥ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

Draw a line through R parallel to ST to use corresponding or alternate interior angles. Using given angles, compute ∠QRS = 40° (example deduction: since 110° and 130° help locate supplementary relationships, the required angle becomes 40°).

4. In Fig. 6.26, if AB ∥ CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Use corresponding/alternate angle properties and supplementary relations. For many figures x = 50° (corresponding) and y = 53° (from 180° − 127° = 53°) depending on labels.

5. In Fig. 6.27, PQ and RS are two mirrors placed parallel. An incident ray AB strikes PQ at B, reflects to BC, strikes RS at C and reflects to CD. Prove AB ∥ CD.

Law of reflection: angle of incidence = angle of reflection. Using parallel mirrors and equal alternate angles repeatedly, the outgoing ray CD is parallel to incoming AB. (Show equal corresponding angles to conclude AB ∥ CD.)

Notes: Exercise 6.2 solutions above outline method and results. For exact numeric answers verify with the labelled diagram (Figs. 6.23–6.27) and substitute values step-by-step.

Summary

Linear pair axiom: adjacent angles formed by a ray on a line sum to 180°, and converse holds.
Vertically opposite angles formed by intersection of two lines are equal.
Lines parallel to the same line are parallel to each other. Use corresponding/alternate angles to prove parallelism.

Use the techniques above to solve exercises: apply linear pair, vertically opposite, corresponding and alternate angle relations, and the law of reflection where relevant.

Lines and Angles — NCERT Solutions (Class 9, Chapter 6)