Question

The slope of any line which is parallel to the x-axis is ______.
A. 0
B. 1
C. -1
D. 2

Answer 1

Hint: The slope of a line $ax+by=c$ , is the ratio of change in the value of y to the change in the value of x. It is represented by m. Its value in this case is $m=-\dfrac{a}{b}$ .
The value of the slope of a line can be anything from 0 to $\pm \infty $.
The slope (m) is also defined as $m=\tan \theta $ , where $\theta $ is the angle made by the line with the positive direction of the x-axis.

Complete step-by-step answer:
A line which is parallel to the x-axis, has the same value of y at any point on it.
Hence, the equation of such a line is $y=k$ , where k is a constant.
The equation can also be written as $0x+by=c$ .
Using the formula $m=-\dfrac{a}{b}$ , we see that $m=0$ .
In other words, there is no change in the value of y for any change in the value of x, therefore, the slope of the line is 0.
It can also be observed from the fact that the angle made by a line parallel to the x-axis with the positive direction of the x-axis is ${{0}^{\circ }}$ , so its slope will be $\tan {{0}^{\circ }}=0$ .
Hence the correct option is A.

Note: The slope of a line passing through two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
The slope of a line parallel to the y-axis is not defined or it is $\infty $ .
A positive value of slope means that the line is rising up whereas a negative slope means that the line is falling, for an increase in the value of x.
Parallel lines have the same slopes.
In other words, if the lines ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$ are parallel, then $\dfrac{{{a}_{1}}}{{{b}_{1}}}=\dfrac{{{a}_{2}}}{{{b}_{2}}}\ \Rightarrow \ {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}=0$ .
Perpendicular lines have the product of their slopes equal to -1.
In other words, if the lines ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$ are perpendicular, then $\dfrac{{{a}_{1}}}{{{b}_{1}}}=-\dfrac{{{b}_{2}}}{{{a}_{2}}}\ \Rightarrow \ {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}=0$ .