## RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS

**Other Exercises**

- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.4
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

**Question 1.
**

**Write the equation representing x-axis.**

**Solution:**

The equation of x-axis is, y = 0.

**Question 2.
**

**Write the equation representing y-axis.**

**Solution:**

The equation of y-axis is, x = 0.

**Question 3.
**

**Write the equation of a line passing through the point (0, 4) and parallel to x-axis.**

**Solution:**

The equation of the line passing through the point (0,4) and parallel to x-axis will be y = 4.

**Question 4.
**

**Write the equation of a line passing through the point (3, 5) and parallel to x-axis.**

**Solution:**

The equation of the line passing through the point (3, 5) and parallel to x-axis will be y = 5.

**Question 5.
**

**Write the equation of a line parallel toy-axis and passing through the point (-3, -7).**

**Solution:**

The equations of the line passing through the point (-3, -7) and parallel to y-axis will be x = -3.

**Question 6.
**

**A line passes through the point (-4, 6) and is parallel to x-axis. Find its equation. A line passes through the point (-4, 6) and is parallel to x-axis. Find its equation.**

**Solution:**

A line parallel to x-axis and passing through the point (-4, 6) will be y = 6.

**Question 7.
**

**Solve the equation 3x – 2 = 2x + 3 and represent the solution on the number line.**

**Solution:**

3x – 2 = 2x + 3

⇒ 3x – 2x = 3 + 2 (By terms formation)

⇒ x = 5

∴ x = 5

Solution on the number line is

**Question 8.
**

**Solve the equation 2y – 1 = y + 1 and represent it graphically on the coordinate plane.**

**Solution:**

2y – 1 = y + 1

⇒ 2y – y = 1 +1

⇒ y = 2

∴ It is a line parallel to x-axis at a distance of 2 units above the x-axis is y = 2.

**Question 9.
**

**If the point (a, 2) lies on the graph of the linear equation 2x – 3y + 8 = 0, find the value of a.**

**Solution:**

∵Points (a, 2) lies on the equation

∵

2x – 3y + 8 = 0

∴ It will satisfy the equation,

Now substituting the value of x = a, y = 2 in the equation

⇒ 2a – 3 x 2+ 8= 0

⇒ 2a + 2= 0

⇒ 2a = -2

⇒ a = \(\frac { -2 }{ 2 }\) = -1

∴ a = -1

**Question 10.
**

**Find the value of k for which the point (1, -2) lies on the graph of the linear equation, x – 2y + k = 0.**

**Solution:**

∵ Point (1, -2) lies on the graph of the equation x – 2y + k = 0

∴ x = 1, y = -2 will satisfy the equation

Now substituting the value of x = 1, y = -2 in it

1-2 (-2) + k = 0

⇒ 1 + 4 + k = 0

⇒ 5+ k = 0 ⇒ k =-5

∴ k = -5

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.