# The graph of y = 6 is a line

a. parallel to x-axis at a distance 6 units from the origin

b. parallel to y-axis at a distance 6 units from the origin

c. making an intercept 6 on the x-axis

d. making an intercept 6 on both the axes

**Solution:**

The given equation is y = 6

It can be written as

y = 0.x + 6

Here y = 6 for every value of x

The points are (0,6), (-1, 6), (1, 6), (3, 6) …

By plotting these points, we obtain a __straight line__ which is __parallel__ to the x-axis at a distance of 6 units from x-axis

Therefore, the line is parallel to the x-axis at a distance 6 units from the origin.

**✦ Try This: **Any solution of the linear equation 5x + 0y + 2 = 0 in two variables is of the form a. (-5/2, m), b. (n, -5/2), c. (0, -5/2), d. (-5, 0)

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 4

**NCERT Exemplar Class 9 Maths Exercise 4.1 Problem 10**

## The graph of y = 6 is a line a. parallel to x-axis at a distance 6 units from the origin, b. parallel to y-axis at a distance 6 units from the origin, c. making an intercept 6 on the x-axis, d. making an intercept 6 on both the axes

**Summary:**

The graph of y = 6 is a line parallel to x-axis at a distance 6 units from the origin

**☛ Related Questions:**

- x = 5, y = 2 is a solution of the linear equation a. x + 2y = 7, b. 5x + 2y = 7, c. x + y = 7, d. 5 . . . .
- If a linear equation has solutions (-2, 2), (0, 0) and (2, -2), then it is of the form a. y - x = 0, . . . .
- The positive solutions of the equation ax + by + c = 0 always lie in the a. 1st quadrant, b. 2nd qua . . . .

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