Equation of Line Class 10 ICSE — Bansal Chapter 14

Written by Greya LakshmiPublished on 28 September 2026

The equation of line class 10 ICSE chapter in Bansal Maths moves coordinate geometry forward from graphs. It defines what makes an equation linear, then builds inclination, slope, and the slope formula through two points. You learn the parallel and perpendicular rules, both intercepts, and the three standard forms of a line’s equation. Each section ends with a worked pattern you can repeat with your own numbers.

Download Equation of a Line Class 10 ICSE Mathematics Chapter 14 PDF

Download the Equation of a Line Class 10 ICSE Mathematics chapter PDF from the Bansal textbook to revise offline or print for practice. It is the full chapter as printed in the book — 19 pages, 13 sections, 74 questions.

What is inside Chapter 14

19 pages · 13 sections · 18 worked examples · 74 exercise questions

Section	Title	Page
14.1	A Basic Concept	p. 3
14.2	Inclination of a line
14.3	Concept of Slope (or, gradient)
14.4	Slope (or, gradient) of a straight line
14.5	The slope of a straight line passing through two given fixed points
14.6	Parallel Lines
14.7	Perpendicular Lines
14.8	Condition for Collinearity of three points
14.9	X-intercept
14.10	Y-intercept
14.11	Equation of a line
14.12	Equally Inclined Lines
14.13	To find the slope and y-intercept of a given line

EXERCISE 14(A): 12 questions
EXERCISE 14(C): 23 questions
EXERCISE 14(E): 39 questions

14.1 A Basic Concept

A linear equation is of first degree in x and y, or in only x or only y. Equations like $3x - 2y = 7$ , $x = 5$ and $y + 3 = 0$ are all linear. Each one graphs as a straight line. In fact, every straight line can be represented by a linear equation.

Two facts connect points with lines. Any point that satisfies the equation of a line lies on that line. Any point through which a line passes always satisfies its equation.

Worked pattern: check whether $(5, -1)$ lies on $3x + 2y = 13$ . Substitute $x = 5$ and $y = -1$ :

$3(5) + 2(-1) = 15 - 2 = 13$

The result is true, so the point lies on the line. Now find $k$ so that $(k, 3)$ lies on $2x + 5y = 19$ . Substitute $x = k$ and $y = 3$ :

$2k + 15 = 19 \Rightarrow k = 2$

EXERCISE 14(A) has 12 questions on these checks. The same substitution method also shows whether a line bisects a segment, by testing the midpoint against the equation. The rest of the Class 10 ICSE Maths resources build on this idea.

14.2 Inclination of a line

The inclination of a line is the angle $\theta$ that the part of the line above the x-axis makes with the x-axis. Direction matters. Measured anticlockwise, $\theta$ is positive. Measured clockwise, $\theta$ is negative.

Two fixed values matter. The x-axis and every line parallel to it have inclination $0^\circ$ . The y-axis and every line parallel to it have inclination $90^\circ$ .

This concept feeds the slope definition in section 14.4. There, the same $\theta$ becomes the tangent of the slope.

14.3 Concept of Slope (or, gradient)

Slope measures how steep a plane or line is. It is the ratio of the vertical rise to the horizontal distance:

$\text{slope} = \frac{\text{vertical rise}}{\text{horizontal distance}} = \frac{AB}{BC}$

This equals $\tan \theta$ , where $\theta$ is the angle the plane makes with the horizontal.

Think of a ramp. The steeper the ramp, the larger the rise for the same horizontal run, so the larger the slope. The ratio, not the actual lengths, decides the steepness.

14.4 Slope (or, gradient) of a straight line

The slope of a straight line is the tangent of its inclination, written as $m$ :

$m = \tan \theta$

The x-axis has inclination $0^\circ$ , so its slope is $\tan 0^\circ = 0$ . The y-axis has inclination $90^\circ$ , so its slope is $\tan 90^\circ = \infty$ , which is not defined.

Sign rules follow the angle. Slope is positive for an acute angle anticlockwise from the x-axis. It is negative for an obtuse angle anticlockwise, or an acute angle clockwise.

Worked pattern: an inclination of $30^\circ$ gives slope $\tan 30^\circ = \frac{1}{\sqrt{3}}$ . An inclination of $135^\circ$ gives slope $\tan 135^\circ = -\tan 45^\circ = -1$ . Angles that are not standard are read from the trigonometric table.

14.5 The slope of a straight line passing through two given fixed points

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two fixed points. The slope of the line through them is:

$m = \frac{y_2 - y_1}{x_2 - x_1} \quad \text{or} \quad m = \frac{y_1 - y_2}{x_1 - x_2}$

Either order works, as long as you keep it consistent. In words, it is the difference of ordinates over the difference of abscissae.

Worked pattern: the line through $(1, 3)$ and $(4, 9)$ has slope:

$\frac{9 - 3}{4 - 1} = \frac{6}{3} = 2$

One warning: a vertical line has $x_1 = x_2$ , so the denominator is zero. Its slope is undefined. The formula simply cannot apply.

14.6 Parallel Lines

Two parallel lines have the same direction, so their inclinations are equal. Equal inclinations mean equal tangents, and equal tangents mean equal slopes:

$\theta = \alpha \Rightarrow \tan \theta = \tan \alpha \Rightarrow m_1 = m_2$

So if two lines are parallel, their slopes are equal. Conversely, if two lines have equal slopes, they are parallel.

Worked pattern: a line with slope 2 is parallel to any other line with slope 2. This rule powers the common exercise of finding an unknown coordinate so that two joins are parallel. Set the two slopes equal and solve.

14.7 Perpendicular Lines

For two perpendicular lines, the product of their slopes is $-1$ :

$m_1 \cdot m_2 = -1$

Each slope is the negative reciprocal of the other. If one slope is $m$ , the perpendicular slope is $-\frac{1}{m}$ .

Why it holds: if one line has inclination $\alpha$ , the other has inclination $\theta = 90^\circ + \alpha$ . Then $\tan \theta = \tan (90^\circ + \alpha) = -\cot \alpha = -\frac{1}{\tan \alpha}$ . So $m_1 \cdot m_2 = -1$ .

Worked pattern: a line of slope 3 has perpendicular slope $-\frac{1}{3}$ , since $3 \times (-\frac{1}{3}) = -1$ . The table below collects the axis values.

Line	Inclination ( $\theta$ )	Slope ( $m$ )
x-axis	$0^\circ$	$0$
y-axis	$90^\circ$	$\infty$ (not defined)
parallel to x-axis	$0^\circ$	$0$
parallel to y-axis	$90^\circ$	$\infty$ (not defined)

14.8 Condition for Collinearity of three points

Three points A, B and C are collinear when they lie on the same straight line. The slope test decides this:

$\text{slope of } AB = \text{slope of } BC$

If the two slopes are equal, the points share one straight path.

Worked pattern: find $a$ so that $A(1, 2)$ , $B(3, 6)$ and $C(4, a)$ are collinear.

$\text{slope } AB = \frac{6-2}{3-1} = 2, \quad \text{slope } BC = \frac{a-6}{4-3} = a-6$

Set $2 = a - 6$ , giving $a = 8$ . The same slope test can show a right-angled triangle or a parallelogram without the distance formula. Compare perpendicular slopes for the triangle, and equal slopes for the parallelogram. This method also connects to the distance and section formulae chapter.

14.9 X-intercept

When a line meets the x-axis at point A, the distance OA from the origin O is the x-intercept. It is the intercept the line makes on the x-axis.

An intercept maps straight to a point. An x-intercept of 5 means the point $(5, 0)$ . An x-intercept of $-8$ means the point $(-8, 0)$ . The y-coordinate is always zero on the x-axis.

14.10 Y-intercept

When a line meets the y-axis at point B, the distance OB from the origin O is the y-intercept.

The mapping mirrors the x-intercept. A y-intercept of 5 means the point $(0, 5)$ . A y-intercept of 6 means the point $(0, 6)$ . The x-coordinate is always zero on the y-axis.

Notice the pattern: an intercept value is the coordinate on that axis, with the other coordinate zero. This value appears directly as $c$ in the slope-intercept form $y = mx + c$ in section 14.11.

14.11 Equation of a line

This is the core section of the chapter. Three standard forms cover almost every question. Choose the form by the data you hold.

Type 1 : (Slope-Intercept Form)

Use this form when the slope and the y-intercept are given. The equation is:

$y = mx + c$

Here $m$ is the slope and $c$ is the y-intercept. Geometrically, $c$ is the ordinate of the point where the line meets the y-axis, the point where $x = 0$ .

Worked pattern: with $m = 2$ and $c = -3$ , the equation is $y = 2x - 3$ . The line cuts the y-axis at $(0, -3)$ and rises 2 units for every 1 unit across.

Type 2 : (Point-Slope Form)

Use this form when the slope and one point on the line are given. The equation is:

$y - y_1 = m(x - x_1)$

The point $(x_1, y_1)$ lies on the line, and $m$ is the slope.

Worked pattern: a line of slope 3 through $(1, 4)$ :

$y - 4 = 3(x - 1) \Rightarrow y = 3x + 1$

Type 3 : (Two-Points Form)

When two points on the line are given, find the slope first, then apply the point-slope form:

$m = \frac{y_2 - y_1}{x_2 - x_1}, \quad y - y_1 = m(x - x_1)$

Worked pattern: the line through $(0, 1)$ and $(2, 5)$ . Slope $= \frac{5-1}{2-0} = 2$ . Then $y - 1 = 2(x - 0)$ , giving $y = 2x + 1$ .

Intercepts convert to points the same way. An x-intercept of 6 gives $(6, 0)$ . A y-intercept of $-9$ gives $(0, -9)$ . You then use the two-points form.

Alternative method: substitute one point into $y = mx + c$ and solve for $m$ . With y-intercept 4 and the point $(3, 10)$ : $10 = 3m + 4$ , so $m = 2$ , and $y = 2x + 4$ .

Here is how the three forms compare:

Form	Given data	Equation	Use it when
Slope-Intercept	slope $m$ , y-intercept $c$	$y = mx + c$	you have both $m$ and $c$
Point-Slope	slope $m$ , point $(x_1, y_1)$	$y - y_1 = m(x - x_1)$	you have $m$ and one point
Two-Points	two points	find $m$ , then point-slope	you have two points

Decision checklist: slope and intercept? Use Type 1. Slope and one point? Use Type 2. Only two points? Use Type 3. When a question gives intercepts, read them as points and apply Type 3.

14.12 Equally Inclined Lines

Equally inclined lines make equal angles with both coordinate axes. A line at $45^\circ$ to the x-axis has slope $\tan 45^\circ = 1$ . A line at $-45^\circ$ has slope $\tan (-45^\circ) = -1$ . So through any given point, there are two equally inclined lines.

Worked pattern: the lines through $(1, 2)$ . With slope 1:

$y - 2 = 1(x - 1) \Rightarrow y = x + 1$

With slope $-1$ :

$y - 2 = -1(x - 1) \Rightarrow x + y = 3$

14.13 To find the slope and y-intercept of a given line

Two steps read the slope and intercept from any line. First, rewrite the equation as $y = mx + c$ . Second, the coefficient of x is the slope, and the constant term with its sign is the y-intercept.

Worked pattern: $4x - 2y - 5 = 0$ .

$-2y = -4x + 5 \Rightarrow y = 2x - \frac{5}{2}$

Slope $= 2$ , y-intercept $= -\frac{5}{2}$ . Carry the sign. The constant $-\frac{5}{2}$ is the intercept, not $+\frac{5}{2}$ .

The same conversion feeds parallel and perpendicular checks in the Class 10 ICSE syllabus. For $4x - 2y = 5$ , rewrite as $y = 2x - \frac{5}{2}$ , so slope $= 2$ . For $2x + ky + 7 = 0$ , rewrite as $y = -\frac{2}{k}x - \frac{7}{k}$ , so slope $= -\frac{2}{k}$ . Set the slopes equal for parallel lines, or multiply to $-1$ for perpendicular lines.

Worked pattern: find the equation of the line through $(4, 1)$ perpendicular to $2x + 5y = 10$ . The given line has slope $-\frac{2}{5}$ , so the perpendicular slope is $\frac{5}{2}$ . Then:

$y - 1 = \frac{5}{2}(x - 4) \Rightarrow 2y - 2 = 5x - 20 \Rightarrow 5x - 2y = 18$

Two mistakes appear often here. First, sign errors when converting to $y = mx + c$ ; the constant keeps its sign. Second, a vertical line has undefined slope, so the two-point formula never applies to it.

What to practise from Equation of Line Class 10 ICSE

This ICSE chapter has three exercise blocks. Work through them in order.

EXERCISE 14(A) — 12 questions on checking whether points lie on a line, finding unknown coefficients, and testing whether a line bisects a join.
EXERCISE 14(C) — 23 questions on writing line equations in the three forms, including intercepts and equally inclined lines.
EXERCISE 14(E) — 39 questions mixing division ratios, medians, altitudes, perpendicular bisectors, rhombus and parallelogram diagonals, and past board questions.

Board questions in this chapter repeat a few shapes. A median joins a vertex to the midpoint of the opposite side. An altitude is perpendicular to the opposite side. A perpendicular bisector passes through the midpoint at right angles. Rhombus diagonals bisect each other at right angles. Practise these four patterns in EXERCISE 14(E) and you cover most past questions. The official CISCE board website lists the Class 10 mathematics syllabus these questions follow.

Equation of Line Class 10 ICSE frequently asked questions

What is the equation of line class 10 ICSE chapter about?

It explains how every straight line has a linear equation and how to write that equation from the data you have. You learn slope, the two intercepts, and the three standard forms of a line’s equation.

What is the slope of a line parallel to the x axis?

Zero. The x-axis has inclination $0^\circ$ , so its slope is $\tan 0^\circ = 0$ . Every line parallel to it has the same slope of 0.

How do I find the equation of a line passing through two points?

Find the slope first using $\frac{y_2 - y_1}{x_2 - x_1}$ . Then apply the point-slope form $y - y_1 = m(x - x_1)$ with one of the points.

What does it mean if the product of the slopes of two lines is minus 1?

The lines are perpendicular. Each slope is the negative reciprocal of the other, like 3 and $-\frac{1}{3}$ .

How do I find the slope and y-intercept of a line like 4x – 2y – 5 = 0?

Rewrite it as $y = mx + c$ . Here $4x - 2y - 5 = 0$ becomes $y = 2x - \frac{5}{2}$ . The slope is 2 and the y-intercept is $-\frac{5}{2}$ .

Reference: ICSE Class 10 Mathematics textbook material, chapter 14: Equation of a Line — CISCE-aligned.

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