Equation of Line Class 10 ICSE — A. Das Gupta Ch 14

Written by Greya LakshmiPublished on 28 September 2026

The equation of line class 10 ICSE chapter in A. Das Gupta’s Mathematics textbook builds the coordinate geometry toolkit for straight lines. It defines inclination and slope, proves the theorems for parallel and perpendicular lines, and derives the three working forms of a line equation. By the end of this chapter, you can write any line’s equation and read its slope and intercepts.

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

Download the Equation of a Straight Line Class 10 ICSE Mathematics chapter PDF from the A. Das Gupta textbook to revise offline or print for practice. It is the full chapter as printed in the book — 30 pages, 4 sections, 64 questions.

What is inside Chapter 14

30 pages · 4 sections · 64 exercise questions

Section	Title
1	Inclination of a Line
2	Slope or Gradient of a Line
3	Some Results on Slope of a Line
6	Equation of An Oblique Line in Various Forms

EXERCISE 14 (A): 27 questions
EXERCISE 14 (B): 37 questions

1 Inclination of a Line

Every straight line on a graph leans at a fixed angle. The inclination $\theta$ of a line is the angle that the part of the line above the $x$ -axis makes with the positive direction of the $x$ -axis, measured anticlockwise. This definition drives everything else in the chapter, because slope is defined from $\theta$ .

Four special values appear again and again. The inclination of the $x$ -axis is $0^{\circ}$ . Every line parallel to the $x$ -axis also has inclination $0^{\circ}$ . The inclination of the $y$ -axis is $90^{\circ}$ , and so is the inclination of every line parallel to the $y$ -axis.

These cases produce the three classifying terms. A horizontal line is any line parallel to the $x$ -axis. A vertical line is any line parallel to the $y$ -axis. An oblique line is parallel to neither axis.

Exam questions rarely ask for $\theta$ on its own. They give an angle and ask for the slope, or give a slope and ask for the angle. Fix the four special values in your memory now, because the next section builds directly on them.

2 Slope or Gradient of a Line

If inclination is the angle, slope is the number that measures it. For a line with inclination $\theta$ , the slope is $m = \tan \theta$ . The book uses slope and gradient as two names for the same value, so a question that asks for the gradient wants the same answer as one that asks for the slope.

Why convert an angle into a number? Angles are awkward in algebra, but a slope like 1 or $-\frac{1}{3}$ slots directly into equations. The sign of the slope also tells you direction: a positive slope rises to the right, and a negative slope falls to the right.

The first questions of Exercise 14 (A) ask you to switch between standard angles and slopes, so these pairs should become automatic.

Inclination $\theta$	Slope $m = \tan \theta$
$0^{\circ}$	0
$30^{\circ}$	$\frac{1}{\sqrt{3}}$
$45^{\circ}$	1
$60^{\circ}$	$\sqrt{3}$

For angles like $38^{\circ}$ or $63^{\circ} 30'$ , use the trigonometric tables to read $\tan \theta$ directly. The reverse move works too. If the slope is 1.732, the tables show $\tan 60^{\circ} = 1.732$ , so $\theta = 60^{\circ}$ .

3 Some Results on Slope of a Line

Every question in the equation of line class 10 ICSE chapter leans on five results. The first two come straight from $m = \tan \theta$ .

The five theorems on slope

The theorems are stated for non-vertical lines, and the reason is built into Theorem 2.

Theorem 1: The slope of the $x$ -axis and of every horizontal line is 0, because their inclination is $0^{\circ}$ and $\tan 0^{\circ} = 0$ .
Theorem 2: The slope of the $y$ -axis and of every vertical line is not defined, because their inclination is $90^{\circ}$ and $\tan 90^{\circ}$ has no value.
Theorem 3: Two non-vertical lines are parallel if and only if their slopes are equal. Equal inclinations force equal tangents, and equal tangents force equal inclinations.
Theorem 4: Two non-vertical lines are perpendicular if and only if the product of their slopes is $-1$ . So the slope of a line perpendicular to AB is $\frac{-1}{\text{slope of AB}}$ .
Theorem 5: The slope of the line through $P(x_1, y_1)$ and $Q(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$ .

Theorem 5 is the workhorse of the chapter. Draw perpendiculars from both points to the $x$ -axis. The vertical and horizontal gaps between the points form a right triangle, and the slope is the ratio of the vertical gap to the horizontal gap. That ratio is exactly $\frac{y_2 - y_1}{x_2 - x_1}$ . The same coordinate differences reappear in the distance and section formulae chapter, where the focus shifts from slope to length and ratio.

For $P(-1, 2)$ and $Q(3, 6)$ :

$m = \frac{6 - 2}{3 - (-1)} = \frac{4}{4} = 1$

This line is parallel to $y = x + 5$ , because that line also has slope 1.

Two slope values cause the most errors in this ICSE chapter: the zero slope of a horizontal line and the undefined slope of a vertical line.

Situation	Common mistake	Correct fact
Horizontal line	Saying its slope is undefined	Slope is 0, since $\tan 0^{\circ} = 0$
Vertical line	Saying its slope is 0	Slope is undefined, since $\tan 90^{\circ}$ has no value
One line vertical, one horizontal	Applying $m_1 m_2 = -1$	The product rule fails; the lines are still perpendicular

Intercepts made by a line on the axes

If a line AB cuts the $x$ -axis at A and the $y$ -axis at B, then OA is the $x$ -intercept and OB is the $y$ -intercept. The signs follow the axes. The $x$ -intercept is positive to the right of the origin and negative to the left. The $y$ -intercept is positive above the origin and negative below it.

A horizontal line never reaches the $x$ -axis, so it has no $x$ -intercept. A vertical line never reaches the $y$ -axis, so it has no $y$ -intercept. Intercept questions in this chapter almost always involve oblique lines for this reason.

Intercepts also give you two points on a line. If the $x$ -intercept is 5 and the $y$ -intercept is 3, the line passes through $(5, 0)$ and $(0, 3)$ , and the two-point form finishes the job.

Equations of the axes and their parallels

Four equations cover the axis and axis-parallel cases. The $x$ -axis is $y = 0$ , because every point on it has ordinate 0. The $y$ -axis is $x = 0$ , because every point on it has abscissa 0. A line parallel to the $y$ -axis at distance $a$ is $x = a$ .

A line parallel to the $x$ -axis at distance $b$ is $y = b$ .

Apply them directly. A line parallel to the $x$ -axis through $(4, -3)$ keeps the $y$ -coordinate fixed, so its equation is $y = -3$ . A line parallel to the $y$ -axis through $(2, 5)$ keeps the $x$ -coordinate fixed, so its equation is $x = 2$ . These four equations appear in questions about rectangles and triangles drawn with axis-parallel sides.

6 Equation of An Oblique Line in Various Forms

This section turns the slope results into equations. An oblique line can be written in three forms, and each form is the fastest route for a particular type of question.

Slope-intercept form

The first form assumes you know the slope $m$ and the $y$ -intercept $c$ . The derivation builds a right triangle on the line. Take a point $P(x, y)$ on the line and draw $PM$ perpendicular to the $x$ -axis. From the $y$ -intercept, drop a perpendicular to $PM$ .

The vertical gap is $y - c$ , the horizontal gap is $x$ , and their ratio is $\tan \theta = m$ . Therefore:

$\frac{y - c}{x} = m$

which rearranges to $y = mx + c$ .

The letters carry geometric meaning. In $y = mx + c$ , the number $m$ is $\tan \theta$ , the tangent of the line’s inclination. The number $c$ is the intercept the line makes on the $y$ -axis. Whenever a question says “find the gradient and the $y$ -intercept of the line”, your first step is to rewrite the equation in this form.

Point-slope form

The second form is for when you know one point and the slope. Start with $y = mx + c$ and substitute the known point $(x_1, y_1)$ . That gives $c = y_1 - mx_1$ . Putting this value back into $y = mx + c$ produces:

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

This form is the fastest when the question hands you a slope and a single point, or when the slope comes from an angle. You substitute once, then rearrange to the form the question wants.

One special case matters. If the line passes through the origin $(0, 0)$ , the form collapses to $y = mx$ . A line through the origin has no separate $y$ -intercept term. The textbook uses this special case in questions that ask for a line through the origin parallel or perpendicular to another line.

Two-point form

The third form needs two points on the line. Since A, P, and B are collinear, the slope of AP equals the slope of AB. Equating the two slope expressions gives:

$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

Here is how the forms connect. Take $A(2, 5)$ and $B(4, 1)$ . The slope is:

$m = \frac{1 - 5}{4 - 2} = \frac{-4}{2} = -2$

Using the point-slope form with $A(2, 5)$ :

$y - 5 = -2(x - 2)$

Expanding gives $y - 5 = -2x + 4$ , so $y = -2x + 9$ . The $y$ -intercept is 9, and the line is now in slope-intercept form. This example shows the three forms are the same line written three ways.

Each form suits a different question type.

Form	Information it needs	Fastest route when
$y = mx + c$	slope $m$ , $y$ -intercept $c$	reading slope and intercept from an equation
$(y - y_1) = m(x - x_1)$	slope and one point	one point plus slope is given
$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$	two points	two points are given
$x = a$ or $y = b$	distance from an axis	line is parallel to an axis

One habit appears throughout the worked solutions in this chapter. Final answers are often rearranged to the form $ax + by + c = 0$ . This is not a separate form, just a tidier way to present the same line.

What to practise from Equation of Line Class 10 ICSE

The equation of line class 10 ICSE chapter has two exercise blocks. EXERCISE 14 (A) holds 27 questions. EXERCISE 14 (B) holds 37 questions.

Exercise 14 (A) drills the foundations. You find slope from inclination, and inclination from slope using the tables. You apply the two-point slope formula, check collinearity through equal slopes, and practise the basic equation forms. Some questions test whether a given point lies on a line, and one asks whether a line bisects a segment.

Exercise 14 (B) raises the difficulty. It mixes parallel and perpendicular checks with medians, altitudes, perpendicular bisectors, circumcentre and orthocentre problems, plus triangle and quadrilateral applications. The question years tagged inside the book (1993, 1994, 1995, 1996, 1997, 2002, 2006, 2007, 2009) show that perpendicular bisector, altitude and median problems recur in board papers.

Work through the exercises in order. The 27 questions of 14 (A) build the slope skills that the 37 questions of 14 (B) turn into geometry applications. For the full set of Class 10 Maths chapters in this collection, see the Class 10 Maths books hub. For the official syllabus and specimen papers, check the CISCE website.

Equation of Line Class 10 ICSE frequently asked questions

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

This ICSE chapter defines the inclination and slope of a line, proves five theorems about slopes, and derives the slope intercept, point slope, and two point forms of a line equation. The two exercises apply these ideas to collinearity, medians, altitudes, and perpendicular bisectors.

How do I find the slope of a line from two points?

Use $m = \frac{y_2 - y_1}{x_2 - x_1}$ . For $P(-1, 2)$ and $Q(3, 6)$ , the slope is $\frac{6 - 2}{3 - (-1)} = 1$ . This is Theorem 5 of Section 3.

Why is the slope of a vertical line not defined?

A vertical line has inclination $90^{\circ}$ , and $\tan 90^{\circ}$ is undefined. The two-point formula also fails because $x_2 - x_1 = 0$ , and division by zero is not possible.

When should I use the point slope form instead of y = mx + c?

Use $(y - y_1) = m(x - x_1)$ when you know one point on the line and its slope but not the $y$ -intercept. Use $y = mx + c$ when the slope and $y$ -intercept are given, or when you want to read them directly from an equation.

How do I check if two lines are parallel or perpendicular?

Rearrange both equations into $y = mx + c$ form. Equal slopes mean the lines are parallel. If the product of the slopes is $-1$ , the lines are perpendicular.

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

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