NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.3 Solutions

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NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.3 Solutions are based on plotting points on a plane if its coordinates are mentioned.

Also check: Important MCQs on Coordinate Geometry

Download PDF: NCERT Solutions for Class 9 Maths Chapter 3 Exercise 3.3 Solutions

Check out NCERT Solutions for Class 9 Maths Chapter 3 Exercise 3.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 3 Coordinate Geometry

Also check other Exercise Solutions of Class 9 Maths Chapter 3 Coordinate Geometry

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Chapter Related Topics
Cartesian System	Quadrant	Distance Formula and Derivation of Coordinate Geometry

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CBSE X Related Questions

1.
Which of the following graphs represents a polynomial with both zeroes being positive?
View Solution
2.
If $PQ$ and $PR$ are tangents to the circle with centre $O$ and radius $4 \text{ cm}$ such that $\angle QPR = 90^{\circ}$, then the length $OP$ is
- $4 \text{ cm}$
- $4\sqrt{2} \text{ cm}$
- $8 \text{ cm}$
- $2\sqrt{2} \text{ cm}$
View Solution
3.
If roots of the quadratic equation $x^2 - k\sqrt{3}x + 2 = 0$ are real and equal, then value of k is
- $-2$
- $\sqrt{\frac{8}{3}}$
- $1$
- $2$
View Solution
4.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is
- $6$
- $4$
- $\pm 4$
- $3$
View Solution
5.
For an acute angle $\theta$, if $\sin \theta = \frac{1}{9}$, then value of $\frac{9 \csc \theta + 1}{9 \csc \theta - 1}$ is
- $0$
- $\frac{80}{81}$
- $1$
- $\frac{82}{80}$
View Solution
6.
If for a data, median is 5 and mode is 4, then mean is equal to :
- 7
- 11
- $\frac{11}{2}$
- $\frac{14}{3}$
View Solution

View All

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NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 3 Coordinate Geometry

CBSE X Related Questions

1.
Which of the following graphs represents a polynomial with both zeroes being positive?

2.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

3.
If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

4.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

5.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

6.
If for a data, median is 5 and mode is 4, then mean is equal to :

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 3 Coordinate Geometry

CBSE X Related Questions

1.Which of the following graphs represents a polynomial with both zeroes being positive?

2.If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

3.If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

4.In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

5.For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

6.If for a data, median is 5 and mode is 4, then mean is equal to :

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Which of the following graphs represents a polynomial with both zeroes being positive?

2.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

3.
If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

4.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

5.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

6.
If for a data, median is 5 and mode is 4, then mean is equal to :