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

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

Explanation

Solution

The slope of a line ax+by=cax+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=abm=-\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=tanθm=\tan \theta , where θ\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=ky=k , where k is a constant.
The equation can also be written as 0x+by=c0x+by=c .
Using the formula m=abm=-\dfrac{a}{b} , we see that m=0m=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{{0}^{\circ }} , so its slope will be tan0=0\tan {{0}^{\circ }}=0 .
Hence the correct option is A.

Note: The slope of a line passing through two points (x1,y1)({{x}_{1}},{{y}_{1}}) and (x2,y2)({{x}_{2}},{{y}_{2}}) is m=y2y1x2x1m=\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 a1x+b1y=c1{{a}_{1}}x+{{b}_{1}}y={{c}_{1}} and a2x+b2y=c2{{a}_{2}}x+{{b}_{2}}y={{c}_{2}} are parallel, then a1b1=a2b2  a1b2a2b1=0\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 a1x+b1y=c1{{a}_{1}}x+{{b}_{1}}y={{c}_{1}} and a2x+b2y=c2{{a}_{2}}x+{{b}_{2}}y={{c}_{2}} are perpendicular, then a1b1=b2a2  a1a2+b1b2=0\dfrac{{{a}_{1}}}{{{b}_{1}}}=-\dfrac{{{b}_{2}}}{{{a}_{2}}}\ \Rightarrow \ {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}=0 .