MathsCBSE Class 11 · Straight Lines

Equations of a Line, Intercepts and Distance from a Point (Class 11)

Updated 17 October 2026 · Fact-checked

Forms of a line: y − y1 = m(x − x1) (point-slope), y = mx + c, (y − y1)/(y2 − y1) = (x − x1)/(x2 − x1) (two points), x/a + y/b = 1 (intercepts a and b). The perpendicular distance of (x1, y1) from Ax + By + C = 0 is |Ax1 + By1 + C|/√(A² + B²). Parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 are |C1 − C2|/√(A² + B²) apart.

Put y=0 and x=0 to get intercepts 4 and 3; the perpendicular from the origin has length |C|/√(A²+B²) = 12/5.

Illustration drawn for QuestionBuddy.

Understand Equations of Lines, Intercepts and Distance from a Point

Standard forms.

Horizontal $y = k$ and vertical $x = h$ .
Point-slope: $y - y_{1} = m (x - x_{1})$ .
Two-point: $y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$ .
Slope-intercept: $y = m x + c$ , where $c$ is the $y$ -intercept.
Intercept form: $\frac{x}{a} + \frac{y}{b} = 1$ , with $x$ -intercept $a$ and $y$ -intercept $b$ (neither zero). To read intercepts from $A x + B y = D$ , divide by $D$ or put $y = 0$ for the $x$ -intercept and $x = 0$ for the $y$ -intercept: for $3 x + 4 y = 12$ they are $4$ and $3$ , and the triangle with the axes has area $\frac{1}{2} ∣ ab ∣ = 6$ .

The general equation is $A x + B y + C = 0$ . The normal form and a deeper treatment of the general equation are formative in 2026-27.

Finding a line. Choose the form that matches the data: a point and a slope (or an angle) use point-slope; two points use two-point; two intercepts use intercept form. For a line parallel to $A x + B y + C = 0$ keep $A, B$ and change $C$ ; for a perpendicular line swap and change the sign: $B x - A y + C^{'} = 0$ .

Intercept conditions. If a line through a given point has intercepts $a, b$ , the point satisfies $\frac{x}{a} + \frac{y}{b} = 1$ ; combine with a second condition (sum of intercepts, area $\frac{1}{2} ab$ ) and solve for $a$ and $b$ .

Distance of a point from a line. The perpendicular distance of $P (x_{1}, y_{1})$ from $A x + B y + C = 0$ is
$ $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}} .$ $
Write the line in the form "=0" first. The distance of the origin is $\frac{∣ C ∣}{A ^{2} + B ^{2}}$ .

Distance between parallel lines. For $A x + B y + C_{1} = 0$ and $A x + B y + C_{2} = 0$ (same $A$ and $B$ — make the coefficients equal first), $d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$ .

Foot and image. The foot $Q$ of the perpendicular from $P$ to the line is $P + s (A, B)$ , where $s$ makes $Q$ lie on the line: $s = - \frac{A x _{1} + B y _{1} + C}{A ^{2} + B ^{2}}$ . The image $P^{'}$ in the line is $P + 2 s (A, B)$ , so $Q$ is the midpoint of $P P^{'}$ .

Lines through an intersection. To find the point where two lines meet, solve the equations simultaneously.

Equations of Lines, Intercepts and Distance from a Point: formulas to remember

Point-slope form: $y - y 1 = m (x - x 1)$; Line through (x1, y1) with slope m.
Intercept form: $x / a + y / b = 1$; Intercepts a on the x-axis, b on the y-axis.
Distance of a point: $d = ∣ A x 1 + B y 1 + C ∣/\sqrt (A^{2} + B^{2})$; Line written as Ax + By + C = 0.
Distance between parallel lines: $d = ∣ C 1 - C 2∣/\sqrt (A^{2} + B^{2})$; Same A and B.
Foot and image: $F oo t : P + s (A, B), s = - (A x 1 + B y 1 + C) / (A^{2} + B^{2}); ima g e : P + 2 s (A, B)$; Foot is the midpoint of P and its image.

How to solve Equations of Lines, Intercepts and Distance from a Point questions

Use these steps to find equations of lines and distances.

1Pick the form that suits the given data (point and slope, two points, intercepts).
2Write the equation, then simplify to $A x + B y + C = 0$ .
3For intercepts, put $x = 0$ and $y = 0$ in turn, or divide by the constant term.
4For distances, write the line as $A x + B y + C = 0$ (and make the coefficients equal for parallel lines), then use the formula with absolute value.
5For a foot or image, find $s$ from the line equation and substitute into $P + s (A, B)$ or $P + 2 s (A, B)$ .
6Check by substituting a known point into the equation.

Equations of Lines, Intercepts and Distance from a Point in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A line as a fare model

Suppose a taxi charges Rs 50 as a fixed fare plus Rs 12 per kilometre, so the fare is $y = 12 x + 50$ for $x$ km.

Identify the slope and the $y$ -intercept, and find the fare for 5 km.

Try it first, then show the solutionSolution

In $y = m x + c$ : slope $m = 12$ (rupees per km) and $y$ -intercept $c = 50$ (the fixed fare).
For $x = 5$ : $y = 12 \times 5 + 50 = 110$ .

Answer: Slope 12; intercept 50; fare Rs 110.

Memory hook: The intercept is the starting value; the slope is the rate.

Distance from a point to a pipe

Suppose a sensor is at $(3, 4)$ and a straight pipe runs along $3 x + 4 y = 12$ (units in metres).

Find the shortest distance from the sensor to the pipe.

Try it first, then show the solutionSolution

Write the line as $3 x + 4 y - 12 = 0$ .
$d = \frac{∣3 \cdot 3 + 4 \cdot 4 - 12∣}{9 + 16} = \frac{∣13∣}{5} = 2.6$ .

Answer: 2.6 m

Memory hook: Substitute the point into the left side, take the modulus, divide by the root of the sum of squares.

Tricks to remember Equations of Lines, Intercepts and Distance from a Point

Short memory aids. Say each one aloud once, then use it on a question the same day.

Intercepts by zeros
x-intercept: set y = 0. y-intercept: set x = 0.
Use it when: Reading intercepts.
Same coefficients first
For two parallel lines, make A and B identical before using |C1 − C2|.
Use it when: Distance between parallel lines.
Swap and negate
Perpendicular to Ax + By + C = 0 is Bx − Ay + k = 0.
Use it when: Perpendicular lines.
Foot is the midpoint
The foot of the perpendicular is the midpoint of the point and its image.
Use it when: Image and foot questions.

Equations of Lines, Intercepts and Distance from a Point: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Point-slope form · 2 marks · Very short answerBoard level

Find the equation of the line through $(3, - 2)$ with slope $\frac{4}{3}$ .

Try it first, then show the solutionSolution

$y + 2 = \frac{4}{3} (x - 3)$ . (1 mark)
$3 y + 6 = 4 x - 12$ , so $4 x - 3 y - 18 = 0$ . (1 mark)

Answer: $4 x - 3 y - 18 = 0$

Example 2: Intercepts and area · 3 marks · Short answerBoard level

Find the intercepts of the line $3 x + 4 y = 12$ on the axes and the area of the triangle it makes with the axes.

Try it first, then show the solutionSolution

Put $y = 0$ : $x = 4$ . Put $x = 0$ : $y = 3$ . (1 mark)
Intercept form: $\frac{x}{4} + \frac{y}{3} = 1$ . (1 mark)
Area $= \frac{1}{2} \times 4 \times 3 = 6$ square units. (1 mark)

Answer: $x$ -intercept 4, $y$ -intercept 3; area 6

Example 3: Two distances · 3 marks · Short answerHarder board level

Find (i) the distance of $(3, - 2)$ from $4 x - 3 y + 5 = 0$ , (ii) the distance between $3 x - 4 y + 7 = 0$ and $3 x - 4 y - 8 = 0$ .

Try it first, then show the solutionSolution

(i) $\frac{∣4 ( 3 ) - 3 ( - 2 ) + 5∣}{16 + 9} = \frac{∣12 + 6 + 5∣}{5} = \frac{23}{5}$ . (1½ marks)
(ii) The coefficients of $x$ and $y$ already match, so $d = \frac{∣7 - ( - 8 ) ∣}{9 + 16} = \frac{15}{5} = 3$ . (1½ marks)

Answer: (i) 4.6 (ii) 3

Example 4: Intercepts and area together · 4 marks · Short answerHarder board level

Find the equation of the line through $(2, 3)$ that makes with the coordinate axes a triangle of area 12 square units in the first quadrant.

Try it first, then show the solutionSolution

Let the intercepts be $a, b > 0$ : $\frac{2}{a} + \frac{3}{b} = 1$ and $\frac{1}{2} ab = 12$ , so $ab = 24$ . (1 mark)
First equation: $2 b + 3 a = ab = 24$ , so $b = \frac{24}{a}$ gives $3 a + \frac{48}{a} = 24$ , i.e. $a^{2} - 8 a + 16 = 0$ . (1 mark)
$(a - 4)^{2} = 0$ , so $a = 4$ and $b = 6$ . (1 mark)
Line: $\frac{x}{4} + \frac{y}{6} = 1$ , i.e. $3 x + 2 y = 12$ . Check: $3 (2) + 2 (3) = 12$ . (1 mark)

Answer: $3 x + 2 y = 12$

Example 5: Foot and image · 5 marks · Long answerHOTS / competency

Find the foot of the perpendicular from $P (3, 4)$ to the line $2 x - y + 1 = 0$ , the distance of $P$ from the line, and the image of $P$ in the line.

Try it first, then show the solutionSolution

Normal direction $(2, - 1)$ . $s = - \frac{2 ( 3 ) - 4 + 1}{4 + 1} = - \frac{3}{5}$ . (1 mark)
Foot $Q = (3 + 2 s, 4 - s) = (\frac{9}{5}, \frac{23}{5})$ . Check on the line: $\frac{18}{5} - \frac{23}{5} + 1 = 0$ . (1½ marks)
Distance $= \frac{∣2 ( 3 ) - 4 + 1∣}{5} = \frac{3}{5}$ (also $∣ s ∣ 5$ ). (1 mark)
Image $P^{'} = P + 2 s (2, - 1) = (3 - \frac{12}{5}, 4 + \frac{6}{5}) = (\frac{3}{5}, \frac{26}{5})$ . (1 mark)
Check: the midpoint of $P$ and $P^{'}$ is $(\frac{9}{5}, \frac{23}{5}) = Q$ . (½ mark)

Answer: Foot $(\frac{9}{5}, \frac{23}{5})$ ; distance $\frac{3}{5}$ ; image $(\frac{3}{5}, \frac{26}{5})$

Where students lose marks in Equations of Lines, Intercepts and Distance from a Point

The slips examiners see most often, and what to write instead.

Mistake: Not writing the line as Ax + By + C = 0
The constant must be on the left with the variables.
Wrong working
Distance of $(3, - 2)$ from $4 x - 3 y = - 5$ using $C = - 5$ .
Right working
Write $4 x - 3 y + 5 = 0$ , then $C = 5$ .
Do this: Move everything to the left side first.
Usually costs: The answer.
Mistake: Using the distance formula for unequal coefficients
Parallel lines must have identical A and B.
Wrong working
$x + 2 y - 3 = 0$ and $2 x + 4 y - 7 = 0$ : $\frac{∣ - 3 + 7∣}{5}$ .
Right working
Divide the second by 2 first.
Do this: Match the coefficients.
Usually costs: The answer.
Mistake: Dropping the modulus
Distance cannot be negative.
Wrong working
$\frac{3 \cdot 1 + 4 \cdot 2 - 20}{5} = - \frac{9}{5}$ .
Right working
$\frac{9}{5}$ .
Do this: Write the bars around the numerator.
Usually costs: 1 mark.
Mistake: Mixing up the intercepts
The x-intercept comes from y = 0.
Wrong working
For $3 x + 4 y = 12$ : "x-intercept 3".
Right working
x-intercept 4, y-intercept 3.
Do this: Substitute zero for the other variable.
Usually costs: The listing mark.

Exam tips

Always write the final equation in the form $A x + B y + C = 0$ unless told otherwise.
Check an equation by substituting the given points.
For distance problems, show the substitution and the modulus.
For foot and image questions, check that the midpoint of $P$ and $P^{'}$ is the foot.
The normal form and the general equation as a topic are formative only in 2026-27; collinearity, family of lines and shifting of origin are not examined.

Practice questions on Equations of Lines, Intercepts and Distance from a Point

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The $x$ -intercept of the line $3 x - 4 y = 12$ is:
(a) 3
(b) 4
(c) $- 4$
(d) 12
Hint
Put $y = 0$ .
Answer
(b). $3 x = 12$ , $x = 4$ .
Q2 · 1 mark · MCQHarder board level
The distance of the point $(1, 2)$ from the line $3 x + 4 y + 5 = 0$ is:
(a) $\frac{16}{5}$
(b) $4$
(c) $3$
(d) $\frac{12}{5}$
Hint
Use $\frac{∣3 + 8 + 5∣}{5}$ .
Answer
(a). $\frac{16}{5} = 3.2$ .
Q3 · 3 marks · Short answerHarder board level
Find the equation of the line through the point of intersection of $x + y = 5$ and $2 x - y = 1$ and perpendicular to $x + 2 y = 3$ .
Hint
Solve the two equations for the point; the perpendicular slope is the negative reciprocal.
Answer
Adding gives $3 x = 6$ , $x = 2$ , $y = 3$ . Slope of $x + 2 y = 3$ is $- \frac{1}{2}$ , so the required slope is $2$ . $y - 3 = 2 (x - 2)$ , i.e. $2 x - y - 1 = 0$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The distance between the parallel lines $x + 2 y = 3$ and $2 x + 4 y = 7$ is $\frac{1}{5}$ .
Reason (R): The distance between $A x + B y + C_{1} = 0$ and $A x + B y + C_{2} = 0$ is $\frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Make the coefficients equal first: divide the second equation by 2.
Answer
(d). The lines are $x + 2 y - 3 = 0$ and $x + 2 y - \frac{7}{2} = 0$ ; $d = \frac{\frac{1}{2}}{5} = \frac{1}{2 5}$ , not $\frac{1}{5}$ . R is correct.
Q5 · 4 marks · Short answerHOTS / competency
Find the equations of the lines parallel to $3 x - 4 y + 2 = 0$ that are at a distance of 2 units from the origin.
Hint
A parallel line has the form $3 x - 4 y + k = 0$ .
Answer
Distance from the origin $= \frac{∣ k ∣}{9 + 16} = \frac{∣ k ∣}{5} = 2$ , so $k = \pm 10$ . The lines are $3 x - 4 y + 10 = 0$ and $3 x - 4 y - 10 = 0$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the intercept form of a line?Show answer
x/a + y/b = 1.
2.What is the distance of (x1, y1) from Ax + By + C = 0?Show answer
|A x1 + B y1 + C| / √(A² + B²).
3.What is the distance between Ax + By + C1 = 0 and Ax + By + C2 = 0?Show answer
|C1 − C2| / √(A² + B²).
4.How do you find the y-intercept?Show answer
Put x = 0.
5.What is the line perpendicular to Ax + By + C = 0?Show answer
Bx − Ay + k = 0.

Equations of Lines, Intercepts and Distance from a Point: frequently asked questions

Which form should I use?

Use point-slope for a point and a slope, two-point for two points, and intercept form for two intercepts.

Why is the origin distance |C|/√(A²+B²)?

Substituting (0, 0) into the distance formula leaves only C in the numerator.

Can a line have a zero intercept?

Yes, a line through the origin has both intercepts zero, and intercept form does not apply to it.

Are the normal form and general equation examined?

The chapter page marks them as formative only in 2026-27.