What is the slope of x-axis?

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Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2024Subjective· 1mImportance★★★★★

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Concept understanding — Slope Calculation

Slope Calculation — From Intuition to Precision

Imagine you're walking up a hill. Some hills are gentle — you barely notice the climb. Others are so steep you have to lean forward and use your hands. That "steepness" is what slope measures. In mathematics, slope tells us how fast a line rises or falls as we move from left to right.

The Intuition: Rise Over Run

Take any two points on a straight line. As you walk from the left point to the right point, two things happen:

You move horizontally — that's the run.
You move vertically — that's the rise (upwards) or fall (downwards).

Slope is simply the ratio:

Slope = (vertical change) ÷ (horizontal change)

If you climb 3 metres while walking 5 metres forward, the slope is $3/5 = 0.6$ . If you descend 2 metres while walking 4 metres forward, the slope is $-2/4 = -0.5$ — negative because you're going downhill.

The Precise Definition

Given two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ on a non-vertical line, the slope $m$ is:

m = \frac{y_2 - y_1}{x_2 - x_1}

The numerator is the rise (change in $y$ ), the denominator is the run (change in $x$ ). The order matters: subtract the first point's coordinates from the second's, consistently.

Watch out

Never divide by zero. If $x_2 = x_1$ , the line is vertical — slope is undefined (not zero, not infinite — just undefined).

What the Number Tells You

Slope value	What the line does
$m > 0$	Rises left to right (uphill)
$m < 0$	Falls left to right (downhill)
$m = 0$	Horizontal (flat)
$m$ undefined	Vertical (straight up/down)

The larger the absolute value $|m|$ , the steeper the line. A slope of $5$ is much steeper than a slope of $0.2$ .

A Worked Example

Find the slope of the line through $(1, 2)$ and $(4, 8)$ .

Step 1: Label the points. Let $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (4, 8)$ .

Step 2: Compute the rise: $y_2 - y_1 = 8 - 2 = 6$ .

Step 3: Compute the run: $x_2 - x_1 = 4 - 1 = 3$ .

Step 4: Divide: $m = \frac{6}{3} = 2$ .

The line rises 2 units vertically for every 1 unit it moves right. …

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The slope of the $x$ -axis is $0$ .

The slope (gradient) of a line is defined as $m = \tan\theta$ , where $\theta$ is the angle the line makes with the positive direction of the $x$ -axis, measured counter-clockwise.

The $x$ -axis itself is inclined at $\theta = 0°$ to the positive $x$ -axis (it lies along it), so

$m = \tan 0° = 0$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$30°$
(a) $-\sqrt{3}$
(b) $-\dfrac{1}{\sqrt{3}}$
(c) $\dfrac{1}{\sqrt{3}}$
(d) $\sqrt{3}$
›Reveal solutionSolution

Convert the angle measured from the $y$ -axis to the equivalent angle measured (anticlockwise) from the $x$ -axis, then take its tangent as the slope.

Slope is defined as $\tan\theta$ where $\theta$ is the angle the line makes with the positive $x$ -axis, measured anticlockwise. The positive $y$ -axis itself is at $90°$ from the positive $x$ -axis. If the line makes a further $30°$ with the positive $y$ -axis (measured anticlockwise, i.e. swinging past the …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The slope of any line parallel to X-axis is:$
(a) $0$
(b) $1$
(c) $-1$
(d) Not defined.
›Reveal solutionSolution

A line parallel to the X-axis is horizontal, so its slope $m=\tan\theta$ with $\theta=0°$ , giving $m=0$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The slope of x-axis is:$
(a) $90^\circ$
(b) 1
(c) 0
(d) $45^\circ$
›Reveal solutionSolution

Slope of a line $= \tan\theta$ , where $\theta$ is the angle it makes with the positive x-axis; the x-axis itself makes a $0^\circ$ angle with itself.

…

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CBSE 2026Set 1A1 mark
$(-4, 3)$
›Reveal solutionSolution

Point-slope form through $(-4,3)$ with slope $\tfrac12$ gives $x-2y+10=0$ .

Using $y-y_1=m(x-x_1)$ with $(x_1,y_1)=(-4,3)$ and $m=\tfrac12$ : …

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CBSE 2025Set ANNUAL1 markMCQ
$(3, -2)$
(a) 0
(b) 1
(c) 4
(d) -1
›Reveal solutionSolution

The slope is $0$ — the line through these points is horizontal.

The slope formula is $m = \dfrac{y_2-y_1}{x_2-x_1}$ .

For $(3,-2)$ and $(7,-2)$ : $m = \dfrac{-2-(-2)}{7-3} = \dfrac{0}{4} = 0$ .

…

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CBSE 2025Set ANNUAL1 mark
$Q. The slope of the line 3x - 4y + 10 = 0 is 4/3. (True/False)$
›Reveal solutionSolution

Writing the line in slope-intercept form gives the true slope, which does not match the statement.

Line: $3x - 4y + 10 = 0$

Solve for $y$ : $4y = 3x+10 \Rightarrow y = \dfrac{3}{4}x + \dfrac{10}{4}$

…

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CBSE 2025Set ANNUAL1 mark
$6x + 3y - 5 = 0$
›Reveal solutionSolution

Rearranging $6x + 3y - 5 = 0$ into $y = mx + c$ form shows the slope $m = -2$ .

Starting from $6x + 3y - 5 = 0$ , solve for $y$ :

$3y = -6x + 5$

$y = -2x + \dfrac{5}{3}$

…

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CBSE 2025Set ANNUAL1 mark
$(3, -2)$
›Reveal solutionSolution

Using the two-point slope formula on $(3,-2)$ and $(-1,4)$ gives slope $-\dfrac{3}{2}$ .

Slope $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ .

With $(x_1,y_1) = (3,-2)$ and $(x_2,y_2) = (-1,4)$ :

…

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CBSE 2024Set ANNUAL1 markMCQ
$30^\circ$
(a) $\dfrac{\sqrt{3}}{2}$
(b) $\sqrt{3}$
(c) $\dfrac{1}{\sqrt{3}}$
(d) $\dfrac{2}{\sqrt{3}}$
›Reveal solutionSolution

The slope is $\tan30^\circ=\dfrac{1}{\sqrt3}$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$(3, -2)$
(a) $0$
(b) $1$
(c) $-1$
(d) Not defined.
›Reveal solutionSolution

The slope of the line through $(3,-2)$ and $(7,-2)$ is $0$ .

Slope formula: $m=\dfrac{y_2-y_1}{x_2-x_1}$ . With $(x_1,y_1)=(3,-2)$ and $(x_2,y_2)=(7,-2)$ :

$m=\dfrac{-2-(-2)}{7-3}=\dfrac{0}{4}=0.$ …

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CBSE 2024Set ANNUAL1 mark
$4x + y + 5 = 0$
›Reveal solutionSolution

Rearranging $4x+y+5=0$ into $y=mx+c$ form shows the slope is $-4$ .

Step 1. Rearrange: $y = -4x - 5$ .

…

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CBSE 2024Set ANNUAL1 mark
$Q. Write the slope of a horizontal line.$
›Reveal solutionSolution

A horizontal line has $y$ constant, so its slope (rate of change of $y$ with $x$ ) is 0.

Step 1. A horizontal line has the form $y = c$ for a constant $c$ .

…

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