**Download chapter wise detailed NCERT Solutions for Class 9 Maths / CBSE Class 9 Maths NCERT Solutions**

**Chapter 1 – Number System**

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**Download chapter wise detailed NCERT Solutions for Class 9 Maths / CBSE Class 9 Maths NCERT Solutions**

**Chapter 1 – Number System**

**Exercise 1.1 – Number System****Exercise 1.2 – Number System****Exercise 1.3 – Number System****Exercise 1.4 – Number System****Exercise 1.5 – Number System****Exercise 1.6 – Number System**

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**Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 – Lines and Angles**

**Q.1 In given figure, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠ SPR = 135° and ∠ PQT = 110°, find ∠ PRQ**

**Given :**

∠ SPR = 135°

∠ PQT = 110°

**To Find :**

∠ PRQ = ?

**Solution:**

∠ SPR + ∠ RPQ = 180° [ Linear Pair of Angles ]

135° + ∠ RPQ = 180°

∠ RPQ = 180° – 135°

∠ RPQ = 45°

∠ PQT + ∠ PQR = 180° [ Linear Pair of Angles ]

110° + ∠ PQR = 180°

∠ PQR = 180° – 110°

∠ PQR = 70°

According to Angle Sum Property, In ∆ PQR

∠ PQR + ∠ PRQ + ∠ RPQ = 180°

70° + ∠ PRQ + 45° = 180°

∠ PRQ + 115° = 180°

∠ PRQ = 180° – 115°

**∠ PRQ = 65°**

**Q. 2 In given figure, ∠ X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of ∆ XYZ, find ∠ OZY and ∠ YOZ**

**Given :**

∠ X = 62°

∠ XYZ = 54°

YO and ZO are bisectors of ∠ XYZ & ∠ XZY

**To Find :**

∠ OZY = ?

∠ YOZ = ?

**Solution:**

According to Angle Sum Property, In ∆ XYZ

∠ XYZ + ∠ YZX + ∠ ZXY = 180°

54° + ∠ YZX + 62° = 180°

∠ YZX + 116° = 180°

∠ YZX = 180° – 116°

∠ YZX = 64°

∠ OZY = 64°/2

**∠ OZY = 32° **[ OZ is angle bisector of ∠ YZX ]

∠ OYZ = 54°/2 = 27° [ OY is angle bisector of ∠ XYZ ]

According to Angle Sum Property, In ∆ OYZ

∠ OYZ + ∠ OZY + ∠ YOZ = 180°

27° + 32° + ∠ YOZ = 180°

59° + ∠ YOZ = 180°

∠ YOZ = 180° – 59°

**∠ YOZ = 21°**

**Q.3 In given figure, if AB || DE, ∠ BAC = 35° and ∠ CDE = 53°, find ∠ DCE.**

**Given :**

AB || DE

∠ BAC = 35°

∠ CDE = 53°

**To Find :**

∠ DCE = ?

**Solution:**

∠ BAC = ∠ CED = 35° [ Alternate interior angles ]

According to Angle Sum Property, In ∆ CDE

∠ CDE + ∠ CED + ∠ DCE = 180°

53° + 35° + ∠ DCE = 180°

88° + ∠ DCE = 180°

∠ DCE = 180° – 88°

∠ DCE = 92°

**Q. 4 In given figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠ TSQ = 75°, find ∠ SQT.**

**Given :**

∠ PRT = 40°

∠ RPT = 95°

∠ TSQ = 75°

**To Find : **∠ SQT = ?

**Solution:**

According to Angle Sum Property, In ∆ PRT

∠ RPT + ∠ PRT + ∠ PTR = 180°

95° + 40° + ∠ PTR = 180°

135° + ∠ PTR = 180°

∠ PTR = 180° – 135°

∠ PTR = 45°

∠ PTR = ∠ STQ = 45° [ Alternate interior angle ]

According to Angle Sum Property, In ∆ STQ

∠ QST + ∠ STQ + ∠ SQT = 180°

75° + 45° + ∠ SQT = 180°

120° + ∠ SQT = 180°

∠ SQT = 180° – 120°

**∠ SQT = 60°**

**Q.5 In given figure, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°, then find the values of x and y.**

**Given :**

PQ ⊥ PS

PQ || SR

∠ SQR = 28°

∠ QRT = 65°

**To Find :**

x = ?

y = ?

**Solution:**

∠ PQR = ∠QRT [ Alternate interior angle ]

x + 28° = 65°

x = 65° – 28°

x = 37°

According to Angle Sum Property, In ∆PQS

x + y + ∠ QPS = 180°

37° + y + 90° = 180°

y + 127° = 180°

y = 180° – 127° **= 53°**

**Q.6 In given figure, the side QR of ∆ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠ QTR = 1/2 ∠ QPR**

**Solution:**

In ∆ QTR, ∠ TRS is an Exterior Angle.

∠ QTR + ∠ TQR = ∠ TRS

∠ QTR = ∠ TRS – ∠ TQR ——– 1

In ∆ PQR, ∠ PRS is an Exterior Angle.

∠ QPR + ∠ PQR = ∠ PRS

∠ QPR + 2 ∠ TQR = 2 ∠ TRS

∠ QPR = 2 ( ∠ TRS – ∠ TQR )

Using Eq 1, ∠ QPR = 2 QTR

**∠ QTR = ****1/2**** ∠ QPR**

**Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 – Lines and Angles**

**Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 – Lines and Angles**

**Q.1 In given figure, find the values of x and y and then show that AB || CD.**

**Solution:**

x + 50˚ = 180˚ [ Linear Pair of Angles ]

x = 180˚ – 50˚

x = 130˚

y = 130˚ [ Vertically Opposite Angles ]

**x = y = 130****˚
**x and y are Alternate interior angle for Line AB and CD

**Q.2 In given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.**

**Given :**

AB || CD

CD || EF

i.e, AB || CD || EF

y : z = 3 : 7

**To Find : **x = ?

**Solution:**

x = z [ Alternate Interior Angles ]

y : z = 3 : 7

Let, y = 3a and z = 7a

x + y = 180˚ [ Co interior Angles ]

z + y = 180˚ [ x = z from above]

7a + 3a = 180˚

10 a = 180˚

a = 180˚/10

a = 18˚

x = z = 7a = 7 x 18˚ = 126˚

**x =126****˚**

**Q.3 In given figure, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE**

**Given :**

AB || CD

EF ⊥ CD

∠ GED = 126°

**To Find :**

∠ AGE = ?

∠ GEF = ?

∠ FGE = ?

**Solution:**

**∠ AGE = ∠ GED = 126° **[ Alternate Interior Angles ]

∠ GED = 126° [ Given ]

∠ GEF + ∠ DEF = 126°

∠ GEF + 90° = 126°

∠ GEF = 126° – 90°

**∠ GEF = 36°
**∠ AGE + ∠ FGE = 180° [ Linear Pair of Angles ]

126° + ∠ FGE = 180°

∠ FGE = 180° – 126°

**Q. 4 In given figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.**

**Given :**

PQ || ST

∠ PQR = 110°

∠ RST = 130°

**To Find : **∠ QRS = ?

**Solution:**

Let us draw XY || ST passing through point R

∠ PQR + ∠ XRQ = 180° [ Co interior Angles ]

110° + ∠ XRQ = 180°

∠ XRQ = 180° – 110° = 70°

∠ XRQ + ∠ QRS = 130° [ Co interior Angles ]

70° + ∠ QRS = 130°

QRS = 130° – 70°

**QRS = 60°**

**Q.5 In given figure, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.**

**Given :**

AB || CD

∠ APQ = 50°

∠ PRD = 127°

**To Find :**

x = ?

y = ?

**Solution:**

∠ APQ = ∠ PQR = 50° [ Alternate interior Angles]

**x = 50°
**∠APR = ∠ PRD = 127° [ Alternate interior Angles]

∠APQ + ∠ QPR = 127°

50° + ∠ QPR = 127°

∠ QPR = 127° – 50°

**Q.6 In given figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD**

**Solution:**

Let us draw, BM ⊥ PQ, CN ⊥ RS

PQ || RS [ Given ]

Therefore, BM || CN

∠ 2 = ∠ 3 [ Alternate interior Angles]

∠ 1 = ∠ 2, ∠ 3 = ∠ 4 [By Laws of Reflection]

∠1 = ∠ 2 = ∠3 = ∠ 4

Also, ∠1 + ∠ 2 = ∠3 + ∠ 4

∠ ABC = ∠ CDB

∠ ABC & ∠ CDB are Alternate interior angles

**AB || CD**

**Download NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2 – Lines and Angles**

**NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 – Lines and Angles**

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

**Given :
**

∠ AOC + ∠ BOE = 70°

∠ BOD = 40°

**To Find :**

∠ BOE = ?

Reflex ∠ COE = ?

**Solution:**

∠ AOC = ∠ BOD [ Vertically opposite angles ]

∠ AOC = ∠ BOD = 40°

∠ AOC = 40°

∠ AOC + ∠ BOE = 70° [Given]

40° + ∠ BOE = 70°

∠ BOE = 70° – 40°

**∠ BOE = 30°**

∠ AOC + ∠ COE + ∠ BOE = 180° [ Angles on Straight line]

∠ AOC + ∠ BOE + ∠ COE= 180°

70° ∠ COE = 180°

∠ COE = 180° – 70°

∠ COE = 110°

Reflex ∠ COE = 360° – 110°

**Reflex ∠ COE = 250°**

** **

**Q.2 In given figure, lines XY and MN intersect at O. If ∠ POY = 90° and a : b = 2 : 3, find c.**

**Given :**

∠ POY = 90°

a : b = 2 : 3

**To Find : **

c = ?

**Solution:**

Let, a = 2x & b = 3x

∠ XOM + ∠ MOP + ∠ POY = 180° [Angle on Straight Line]

b + c + 90° = 180°

3x + 2x + 90° = 180°

3x + 2x = 180° – 90°

5x = 90°

x = (90°)/5

x = 18°

a = 2x = 2 x 18° = 36°

b = 3x = 3 x 18° = 54°

∠ XOM + ∠ XON = 180° [Linear Pair]

54° + c = 180°

c = 180° – 54°

**c = 126°**

**Q.3 In given figure, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.**

**Given :**

∠ PQR = ∠ PRQ

**To Prove :**

∠ PQS = ∠ PRT

**Solution:**

∠ PQS + ∠ PQR = 180° [ Linear Pair ]

∠ PQR = 180° – ∠ PQS ——— 1

∠ PRT + ∠ PRQ = 180° [ Linear Pair ]

∠ PRQ = 180° – ∠ PRT ———- 2

∠ PQR = ∠ PRQ [ Given ]

On Equating Equation 1 and 2, we get

180° – ∠ PQS = 180° – ∠ PRT

**∠ PQS = ∠ PRT**

**Q.4 In given figure, if x + y = w + z, then prove that AOB is a line.**

**Given : **x + y = w + z

**To Prove : **AOB is a line

**Proof:**

x + y + w + z = 360° [ Complete Angle ]

x + y = w + z [ Given ]

Since, x + y + x + y = 360°

2 ( x + y ) = 360°

( x + y ) = (360°)/2

( x + y ) = 180°

**Since, x and y form a Linear Pair, AOB is a line**

**Q. 5 In given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ ROS = (∠ QOS – ∠ POS).**

**Given :**

OR is perpendicular to line PQ

i.e, ∠ POR = ∠ QOR = 90°

**To Prove :**

∠ ROS = (∠ QOS – ∠ POS).

**Proof:**

∠ POR = 90° [Given : OR is perpendicular to line PQ]

∠ POS + ∠ ROS = 90°

∠ ROS = 90° – ∠ POS ———– 1

∠ QOR = 90° [Given : OR is perpendicular to line PQ]

∠ QOS – ∠ ROS = 90°

∠ ROS = ∠ QOS – 90° ———– 2

∠ 2ROS = ∠ QOS – ∠POS [On adding Eq 1 and 2]

∠ ROS = (∠ QOS – ∠ POS)

**Q.6 It is given that ∠ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠ QYP**

**Given :**

∠ XYZ = 64°

YQ bisects ∠ ZYP

i.e, ∠ ZYQ = ∠ QYP

**To Find:**

∠ XYQ = ?

Reflex ∠ QYP = ?

**Solution:**

∠ XYZ + ∠ ZYQ + QYP = 180° [Angles on a Straight line]

∠ XYZ + 2 ∠ QYP = 180° [∠ ZYQ = ∠ QYP]

64° + 2 ∠ QYP = 180°

2 ∠ QYP = 180° – 64°

2 ∠ QYP = 116°

∠ QYP = 58°

∠ ZYQ = ∠ QYP = 58°

**Reflex ∠ QYP = 360° – 58° = 302°**

∠ XYQ = ∠ XYZ + ∠ ZYQ

∠ XYQ = 64° + 58°

**∠ XYQ = 122°**

**NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 – Lines and Angles**

**Download chapter wise detailed Maths NCERT Solutions for Class 6, Class 7, Class 8, Class 9 and Class 10 **

**Chapter 1 – Real Numbers**

**Exercise 1.1 – Real Numbers****Exercise 1.2 – Real Numbers****Exercise 1.3 – Real Numbers****Exercise 1.4 – Real Numbers**

**Chapter 2 – Polynomials**

**Chapter 3 – Pair of Linear Equations in two Variables**

**Exercise 3.3 – Pair of Linear Equations in two Variables****Exercise 3.4 – Pair of Linear Equations in two Variables****Exercise 3.5 – Pair of Linear Equations in two Variables****Exercise 3.6 – Pair of Linear Equations in two Variables**

**Chapter 4 – Quadratic Equations**

**Exercise 4.1 – Quadratic Equations****Exercise 4.2- Quadratic Equations****Exercise 4.3- Quadratic Equations****Exercise 4.4- Quadratic Equations**

**Chapter 5 – Arithmetic Progressions**

**Exercise 5.1 – Arithmetic Progressions****Exercise 5.2 – Arithmetic Progressions****Exercise 5.3 – Arithmetic Progressions**

**Chapter 6 – Triangles**

**Chapter 7 – Coordinate Geometry **

**Chapter 8 – Introduction to Trignometry**

**Exercise 8.1 – Introduction to Trignometry****Exercise 8.2 – Introduction to Trignometry****Exercise 8.3 – Introduction to Trignometry****Exercise 8.4 – Introduction to Trignometry**

**Chapter 9 – Some Applications of Trignometry**

**Chapter 10 – Circles**

**Chapter 11 – Constructions**

**Chapter 12 – Area related to Circles**

**Exercise 12.1 – Area related to Circles****Exercise 12.2 – Area related to Circles****Exercise 12.3 – Area related to Circles**

**Chapter 1 – Number System**

**Exercise 1.1 – Number System****Exercise 1.2 – Number System****Exercise 1.3 – Number System****Exercise 1.4 – Number System****Exercise 1.5 – Number System****Exercise 1.6 – Number System**

**Chapter 2 – Polynomials**

**Exercise 2.1 – Polynomials****Exercise 2.2 – Polynomials****Exercise 2.3 – Polynomials****Exercise 2.4 – Polynomials****Exercise 2.5 – Polynomials**

**Chapter 3 – Coordinate Geometry**

**Chapter 4 – Linear Equations in two Variables**

**Exercise 4.1 – Linear Equations in two Variables****Exercise 4.2 – Linear Equations in two Variables****Exercise 4.3 – Linear Equations in two Variables****Exercise 4.4 – Linear Equations in two Variables**

**Chapter 5 – Introduction to Euclid’s Geometry**

**Chapter 6 – Lines And Angles**

**Chapter 8 – Quadrilaterals**

**Chapter 9 – Areas of Parallelograms And Triangles**

**Exercise 9.1 – Areas of Parallelograms And Triangles****Exercise 9.2 – Areas of Parallelograms And Triangles****Exercise 9.3 – Areas of Parallelograms And Triangles**

**Chapter 10 – Circles**

**Exercise 10.1 – Circles****Exercise 10.2 – Circles****Exercise 10.3 – Circles****Exercise 10.4 – Circles****Exercise 10.5 – Circles**

**Chapter 12 – Heron’s Formula**

**Chapter 15 – Probability**

**Chapter 1 – Rational Numbers**

**Chapter 2 – Linear Equation in one Variable**

**Exercise 2.1 – Linear Equation in one Variable****Exercise 2.2 – Linear Equation in one Variable****Exercise 2.3 – Linear Equation in one Variable****Exercise 2.4 – Linear Equation in one Variable****Exercise 2.5 – Linear Equation in one Variable****Exercise 2.6 – Linear Equation in one Variable**

**Chapter 3 – Understanding Quadrilaterals**

**Exercise 3.1 – Understanding Quadrilaterals****Exercise 3.2 – Understanding Quadrilaterals****Exercise 3.3 – Understanding Quadrilaterals****Exercise 3.4 – Understanding Quadrilaterals**

**Chapter 4 – Practical Geometry**

**Exercise 4.1 – Practical Geometry****Exercise 4.2 – Practical Geometry****Exercise 4.3 – Practical Geometry****Exercise 4.4 – Practical Geometry****Exercise 4.5 – Practical Geometry**

**Chapter 5 – Data Handling**

**Chapter 6 – Square and Square roots**

**Exercise 6.1 – Square and Square roots****Exercise 6.2 – Square and Square roots****Exercise 6.3 – Square and Square roots****Exercise 6.4 – Square and Square roots**

**Chapter 7 – Cube and Cube roots**

**Chapter 8 – Comparing Quantities**

**Exercise 8.1 – Comparing Quantities****Exercise 8.2 – Comparing Quantities****Exercise 8.3 – Comparing Quantities**

**Chapter 9 – Algebraic Expressions and Identities**

**Exercise 9.1 – Algebraic Expressions and Identities****Exercise 9.2 – Algebraic Expressions and Identities****Exercise 9.3 – Algebraic Expressions and Identities**

**Chapter 11 – Mensuration**

**Exercise 11.1 – Mensuration****Exercise 11.2 – Mensuration****Exercise 11.3 – Mensuration****Exercise 11.4 – Mensuration**

**Chapter 12 Exercise 12.1 – Exponents and Powers**

**Chapter 13 – Direct and Inverse Proportions**

**Chapter 14 – Factorisation**

**Chapter 15 – Introduction to Graphs**

**Exercise 15.1 – Introduction to Graphs****Exercise 15.2 – Introduction to Graphs****Exercise 15.3 – Introduction to Graphs**

**Chapter 1 – Integers****Chapter 2 – Fractions and Decimals****Chapter 3 – Data Handling****Chapter 4 – Simple Equations****Chapter 5 – Lines and Angles****Chapter 6 – Triangle and its properties****Chapter 7 – Congruence of Triangles****Chapter 8 – Comparing Quantities****Chapter 9 – Rational Numbers****Chapter 10 – Practical Geometry****Chapter 11 – Perimeter and Area****Chapter 12 – Algebraic Expressions**

**Chapter 1 – Knowing our Numbers****Chapter 2 – Whole Numbers****Chapter 3 – Playing with Numbers****Chapter 4 – Basic Geometrical Ideas****Chapter 5 – Understanding Elementary Shapes****Chapter 6 – Integers****Chapter 8 – Decimals****Chapter 9 – Data Handling****Chapter 10 – Mensuration****Chapter 11 – Algebra****Chapter 12 – Ratio and Proportion**

Maths Class 6 Maths Class 7 Maths Class 8

Coordinate geometry is one of the most important and exciting ideas of mathematics. In particular it is central to the mathematics students meet at school. It provides a connection between algebra and geometry through graphs of lines and curves.

A Cartesian coordinate system is a coordinate system that specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed distances from the point to two fixed perpendicular directed lines, measured in the same unit of length.

Here +x indicates about positive x-axis and –x indicates negative x-axis.

Similarly +y indicates positive y-axis and –y indicates negative y-axis.

You have observed that the axes have divided the plane into four parts. These four parts are called as quadrants which are numbered as I, II, III and IV anticlockwise from +x. So the plane consists of the axes and quadrants. We call the plane, the Cartesian plane or the coordinate plane or xy plane and the axis are called as coordinate axis.

**Example 1**

In which quadrant does the point P(-2, 3) lie?

**Solution**

Coordinate axes divide the plane of the graph paper into four quadrants.

In Quadrant I, sign of coordinates are (+, +)

In Quadrant II, sign of coordinates are (-, +)

In Quadrant III, sign of coordinates are (-, -)

In Quadrant IV, sign of coordinates are (+, -)

Since, In Point P (-2, 3) x-coordinate is negative while y-coordinate is positive.

Hence, Point P (-2, 3) lie in Quadrant II.

**Example 2**

In which quadrant does the point Q (3, -4) lie?

**Solution**

Coordinate axes divide the plane of the graph paper into four quadrants.

In Quadrant I, sign of coordinates are (+, +)

In Quadrant II sign of coordinates are (-, +)

In Quadrant III sign of coordinates are (-, -)

In Quadrant IV sign of coordinates are (+, -)

Since, In Point Q (3, -4) x-coordinate is positive while y-coordinate is negative.

Hence, Point Q (3, -4) lie in Quadrant IV.

**Example 3**

In which quadrant does the point O (-3, -2) lie?

**Solution**

Coordinate axes divide the plane of the graph paper into four quadrants.

In Quadrant I, sign of coordinates are (+, +)

In Quadrant II sign of coordinates are (-, +)

In Quadrant III sign of coordinates are (-, -)

In Quadrant IV sign of coordinates are (+, -)

Since, In Point O (-3, -2) both x and y coordinate are negative.

Hence, Point O (-3, -2) lie in Quadrant III.

**Example 4**

In which quadrant does the point P(1, 4) lie?

**Solution**

Coordinate axes divide the plane of the graph paper into four quadrants.

In Quadrant I, sign of coordinates are (+, +)

In Quadrant II, sign of coordinates are (-, +)

In Quadrant III, sign of coordinates are (-, -)

In Quadrant IV, sign of coordinates are (+, -)

Since, In Point P (1, 4) both x and y coordinate are positive

Hence, Point P (1, 4) lie in Quadrant I.

We write the coordinates of a point using the following instruction.

1) The x-coordinates of a point is its perpendicular distance from y-axis measured along x-axis (Positive along the positive x-axis and Negative along the negative x-axis). For the point P it is 3. The **X-****coordinate is also called as the abscissa**.

2) The y coordinates of a point is its perpendicular distance from x-axis measured along the y axis (Positive along the positive y-axis and negative along the negative y-axis). For the point P that will be 4. **Y-coordinate is also known as the ordinate**.

3) In stating the coordinates of a point in the coordinate plane, the x-coordinate comes first and then the y-coordinate. We place the coordinates in the bracket i.e, **(3,4)**

**Example 1**

Write the coordinates of point P.

**Solution**

The point P is at 1 unit distance from y-axis and 3 units distance from x-axis. Therefore point P is (1,3)

**Example 2**

Write the coordinates of point B.

**Solution**

Point is at 2 unit distance from y-axis along positive x-axis so the x coordinate is 2 and from x-axis along negative y-axis the distance is 2 so the y-coordinate is -2. So the point B will be (2, -2).

**Example 3**

Write the coordinates of point Q.

**Solution**

The point is on y-axis so the distance from y-axis along the x-axis will be zero because it is on the y-axis. So the x coordinate will be zero. Similarly, the distance from x-axis along positive y-axis is 2 so the y-coordinate will be 2. So point Q is (0,2)

Let the coordinates of a point M be (-3, 2). We want to plot this point in the coordinate plane. We draw the coordinate axis and choose our units such that one centimeter represents one unit on both the axes. The coordinates of the point (-3, 2)tell us that the distance of this point from y axis along negative x-axis is 3 units and the distance of the point from the x-axis along the positive y-axis is 2 units. Starting from the origin we count 3 units on the negative x-axis and mark the corresponding point. Now starting from this point we move towards positive y axis and count 2 unit distance and mark the point. This point will be (-3, 2).

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