Straight Lines: Class 11 Maths · Chapter 9

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

75 min
1 simulation
75 practice questions
20 MCQs

What you will learn in Straight Lines

Recall and apply the distance, section, midpoint and triangle-area formulas, including the zero-area test for collinear points.
Find the inclination and slope of a line from an angle or from two of its points, and recognise horizontal and vertical lines.
Decide from slopes whether two lines are parallel or perpendicular and whether three points are collinear.
Find the acute angle between two lines from their slopes, and a missing slope from a given angle.
Write the equation of a line in point-slope, two-point, slope-intercept, intercept and general form, and convert between the forms.
Find the distance of a point from a line and the distance between two parallel lines.
Solve problems on concurrent lines, images in a line, distances along a direction and loci that are straight lines.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 9, pp. 151–175. Not affiliated with or endorsed by NCERT.

9.1 Introduction

Draw a straight line on graph paper and you have a picture. Write $y = 2 x + 1$ and you have the same object as a sentence of algebra. Coordinate geometry is the art of switching between the two: each point becomes a pair of numbers, each curve becomes an equation, and many geometric questions can then be settled by calculation.

René Descartes laid out this programme in La Géométrie (1637), and Pierre de Fermat reached similar ideas independently in the same years. Studying curves through their equations is now called analytical geometry. In earlier classes you plotted points and found distances, section points and areas. This chapter turns those tools on the simplest curve of all, the straight line, and the number that does most of the work is its slope.

Tools from earlier classes

The point $(x, y)$ lies $∣ x ∣$ units from the $y$ -axis and $∣ y ∣$ units from the $x$ -axis, and the signs tell you which side. So $(- 5, 2)$ is 5 units left of the $y$ -axis and 2 units above the $x$ -axis, while $(0, - 3)$ sits on the $y$ -axis itself. Fig. 9.1 runs through the four results we keep using.

2/4A right triangle with legs 8 and 6 gives $A B = 10$ .

Fig. 9.1One pair of points, four old formulas.NCERT Fig. 9.1 · p. 151

A coordinate grid from −4 to 6 across and −3 to 5 up. Step 1 plots A(−3, −2) and B(5, 4) and joins them with a blue segment. Step 2 adds a horizontal leg of length 8 and a vertical leg of length 6 meeting at a right angle at (5, −2), so AB = 10. Step 3 marks the midpoint M(1, 1) with two equal tick marks on AB. Step 4 shades triangle ABC with C(5, −2) and gives its area as half of 6 times 8, which is 24.

Distance. Join $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ and complete a right triangle whose legs are parallel to the axes. The legs are $∣ x_{2} - x_{1} ∣$ and $∣ y_{2} - y_{1} ∣$ , so Pythagoras' theorem gives

distance formula

\begin{aligned} P Q^{2} & = (x_{2} - x_{1})^{2} \\ + (y_{2} - y_{1})^{2} \end{aligned}

(9.1)

For $(1, - 2)$ and $(4, 2)$ we get $P Q^{2} = 9 + 16 = 25$ , so $P Q = 5$ .

Section formula. The point $R$ that divides $P Q$ internally in the ratio $m : n$ , counted from $P$ , has

section formula

\begin{aligned} x & = \frac{m x_{2} + n x_{1}}{m + n} \\ y & = \frac{m y_{2} + n y_{1}}{m + n} \end{aligned}

(9.2)

For example, the point dividing the join of $(2, 1)$ and $(8, 10)$ in the ratio $1 : 2$ is $(\frac{8 + 4}{3}, \frac{10 + 2}{3}) = (4, 4)$ . With $m = n$ the formula gives the midpoint $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

Area of a triangle. For vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ , the area $Δ$ is

area of a triangle

\begin{aligned} Δ = \frac{1}{2} ∣ & x_{1} (y_{2} - y_{3}) \\ + & x_{2} (y_{3} - y_{1}) \\ + & x_{3} (y_{1} - y_{2}) ∣ \end{aligned}

(9.3)

The modulus keeps the area positive in whatever order you list the vertices. For $(1, 1)$ , $(4, 3)$ and $(2, 6)$ the bracket is $- 3 + 20 - 4 = 13$ , so the area is $6.5$ square units. If the bracket is 0, the triangle has collapsed: the three points are collinear.

SimulationDistances and collinear points

A coordinate grid from −6 to 6 on both axes with three draggable points A(−4, −2), B(2, 1) and C(4, 2), each showing its coordinates and snapping to whole numbers. The distances AB, BC and AC are drawn and measured. At the start the three points lie on one line, so AB + BC equals AC; moving any point off that line makes AB + BC larger than AC.

Fig. 9.2Drag the points and watch the three distances. When

B

lies on segment

A C

, the two short distances add up to the long one.

Try this

Check that $A B + B C = A C$ at the start. Now drag $C$ one square up. What happens to the sum?
Keep $A$ at $(- 4, - 2)$ and move $B$ until $A B = 5$ . Which right triangle did you use?
Put $B$ at the midpoint of $A C$ . Do its coordinates agree with the midpoint formula?

Quick check

1 of 3

Find the distance between

(- 2, 3)

and

(4, - 5)

Show answer

Answer 10

The legs are

∣ 4 - (- 2) ∣ = 6

and

∣ - 5 - 3 ∣ = 8

, so the distance is

\sqrt{36 + 64} = 10

by (9.1).

9.2 Slope of a Line

Two straight roads can feel very different: one climbs gently, another climbs hard, a third runs downhill. One number captures this tilt. We build it from the angle between the line and the $x$ -axis.

Measure the tilt with an angle. Stand where the line crosses the $x$ -axis, face the positive $x$ -direction, and turn anticlockwise until you face along the line. The angle you turned through, $θ$ , is the inclination of the line. Turning a full straight angle would only bring you back onto the same line, so $0^{\circ} \leq θ < 180^{\circ}$ . A horizontal line needs no turn ( $θ = 0^{\circ}$ ) and a vertical line needs a quarter turn ( $θ = 90^{\circ}$ ).

Definition

Slope

If a line has inclination $θ$ and $θ \neq 90^{\circ}$ , its slope (or gradient) is $m = \tan θ$ . A vertical line has no slope.

ExampleA line with inclination $45^{\circ}$ has slope $\tan 45^{\circ} = 1$ .

slope

m = \tan θ, θ \neq 90^{\circ}

(9.4)

The sign of the slope (ढाल) tells you the direction of travel as you move from left to right. A positive slope means the line climbs ( $θ$ acute), a negative slope means it falls ( $θ$ obtuse), and slope 0 means it stays level. So the $x$ -axis has slope 0, while the $y$ -axis has no slope at all because $\tan 90^{\circ}$ is undefined. Fig. 9.3 turns one line through all four cases.

2/4 $θ = 125^{\circ}$ : the line falls and $m < 0$ .

Fig. 9.3Inclination is measured anticlockwise from the positive

x

-axis.NCERT Fig. 9.2 · p. 153

Axes with a blue line through the point K(1, 0). Step 1: the line makes an acute angle of 35 degrees with the positive x-axis, marked by an arc, and its slope tan 35° is about 0.70. Step 2: the line turns to 125 degrees, the arc now spans an obtuse angle, and the slope is about −1.43. Step 3: the line lies along the x-axis, inclination 0 and slope 0. Step 4: the line stands vertical with a right-angle mark; tan 90° is undefined, so the line has no slope.

9.2.1 Slope from the coordinates of two points

Two points fix a line, so they must also fix its slope. Take $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ on a non-vertical line, so that $x_{1} \neq x_{2}$ . Going from $P$ to $Q$ , call the horizontal change $x_{2} - x_{1}$ the run and the vertical change $y_{2} - y_{1}$ the rise (Fig. 9.4).

DerivationSlope as rise over run

The line climbs ( $θ$ acute). Put $Q$ to the right of $P$ and above it, and let $M (x_{2}, y_{1})$ complete the right triangle $P M Q$ . Since $P M$ is parallel to the $x$ -axis, the angle of the triangle at $P$ equals $θ$ (corresponding angles). So

\begin{aligned} \tan θ & = \frac{M Q}{P M} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}

The line falls ( $θ$ obtuse). Now $Q$ is to the right of $P$ and below it. The angle of triangle $P M Q$ at $P$ is $180^{\circ} - θ$ and the vertical side is $M Q = y_{1} - y_{2}$ . Using $\tan (180^{\circ} - θ) = - \tan θ$ ,

\begin{aligned} \tan θ & = - \frac{y_{1} - y_{2}}{x_{2} - x_{1}} \\ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}

Both cases give the same rule. Renaming $P$ as $Q$ and $Q$ as $P$ flips the sign of both differences, so the ratio is unchanged.

slope from two points

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

(9.5)

where

$x_{1} \neq x_{2}$: the line is not vertical

1/3Rise 3 over run 5: $m = \frac{3}{5}$ .

Fig. 9.4Slope is rise divided by run, whichever point you start from.NCERT Fig. 9.3 · p. 153

Step 1: a blue line climbs through P(−2, −1) and Q(3, 2); a right triangle under it has a horizontal run labelled x2 − x1 and a vertical rise labelled y2 − y1, giving slope 3/5. Step 2: a dashed line falls from P(−2, 2) to Q(3, −2); the run is 5 and an arrow on the vertical side points down for a rise of −4, so the slope is −4/5. Step 3: the names P and Q are swapped and the arrow points up; now the run is −5 and the rise is 4, and the slope is still −4/5.

9.2.2 Conditions for parallel and perpendicular lines

Take two lines $l_{1}$ and $l_{2}$ , neither of them vertical. Call their slopes $m_{1}$ , $m_{2}$ and their inclinations $α$ , $β$ .

PropertyParallel lines

Two non-vertical lines are parallel exactly when their slopes are equal: $m_{1} = m_{2}$ .

PropertyPerpendicular lines

Two non-vertical lines are perpendicular exactly when $m_{1} m_{2} = - 1$ , that is, when $m_{2} = - \frac{1}{m_{1}}$ .

Suppose $l_{1} ⊥ l_{2}$ with $β > α$ . In the triangle the two lines make with the $x$ -axis, $β$ is an exterior angle, so $β = α + 90^{\circ}$ . Since $\tan (α + 90^{\circ})$ equals $- \cot α$ , we get $m_{2} = - \frac{1}{m_{1}}$ , that is, $m_{1} m_{2} = - 1$ .

Conversely, suppose $m_{1} m_{2} = - 1$ . Then $m_{2} = - \cot α$ , which is the tangent of $α + 90^{\circ}$ and of $α - 90^{\circ}$ ; one of these lies in $[0^{\circ}, 180^{\circ})$ and must be $β$ . So the inclinations differ by $90^{\circ}$ and the lines are perpendicular.

Fig. 9.5 shows where the $- 1$ comes from. Turn a run-and-rise triangle through $90^{\circ}$ : the run becomes the rise, the rise becomes a run in the opposite direction, and the slope $\frac{2}{3}$ becomes $- \frac{3}{2}$ . Slopes of perpendicular lines are negative reciprocals.

3/3 $\frac{2}{3} \times (- \frac{3}{2}) = - 1$ .

Fig. 9.5Equal slopes mean parallel; slopes with product

- 1

mean perpendicular.NCERT Fig. 9.5 · p. 155

Step 1: two parallel lines, solid and dashed, each carries a slope triangle with run 3 and rise 2, so both slopes are 2/3. Step 2: a dotted line through the origin at right angles to the solid line appears, and the triangle on the solid line turns 90 degrees anticlockwise about the origin onto it; its run of 3 becomes a rise of 3 and its rise of 2 becomes a run of 2 to the left, giving slope −3/2. Step 3: a right-angle mark at the origin and the product (2/3)(−3/2) = −1.

Collinear points. Points $A$ , $B$ and $C$ lie on one line exactly when the slope of $A B$ equals the slope of $B C$ (or both segments are vertical). Equal slopes make the segments parallel, and since they share the point $B$ they lie on a single line. This test is often quicker than the area test of (9.3).

Example 9.1NCERT Example 1 · p. 155

Four slopes

Work out the slope, where it exists, of each line: (a) through $(1, - 3)$ and $(5, 7)$ ; (b) through $(1, - 3)$ and $(- 6, - 3)$ ; (c) through $(1, - 3)$ and $(1, 8)$ ; (d) with inclination $150^{\circ}$ .

Try it yourself first. Then reveal the solution one step at a time.

9.2.3 Angle between two lines

Where two lines cross, four angles appear but only two sizes: opposite angles are equal and neighbouring ones are supplementary. So a single angle $θ$ settles everything, and its neighbour is $ϕ = 180^{\circ} - θ$ .

DerivationThe angle formula

Let $L_{1}$ and $L_{2}$ have slopes $m_{1} = \tan α_{1}$ and $m_{2} = \tan α_{2}$ , with $α_{1} < α_{2}$ , neither equal to $90^{\circ}$ (Fig. 9.6). In the triangle the lines cut off with the $x$ -axis, $α_{2}$ is an exterior angle. So $α_{2} = α_{1} + θ$ , where $θ$ is the angle of that triangle at the crossing point, and $\tan θ = \tan (α_{2} - α_{1})$ . Expand with the formula for $\tan (A - B)$ and put $\tan α_{1} = m_{1}$ , $\tan α_{2} = m_{2}$ :

\tan θ = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}

This needs $1 + m_{1} m_{2} \neq 0$ ; if it is 0, the lines are perpendicular. The neighbouring angle has $\tan ϕ = \tan (180^{\circ} - θ) = - \tan θ$ . One of $\tan θ$ and $\tan ϕ$ is positive, and the angle with the positive tangent is the acute one. Taking the modulus picks it out.

acute angle between two lines

\tan θ = ∣ \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} ∣

(9.6)

where

$θ$: the acute angle; the obtuse angle is $180^{\circ} - θ$
$1 + m_{1} m_{2}$: non-zero (otherwise the lines are perpendicular)

2/3 $θ = α_{2} - α_{1}$ ; here $\tan θ = \frac{3 - 1 / 2}{1 + 3 / 2} = 1$ .

Fig. 9.6The exterior-angle fact gives

θ = α_{2} - α_{1}

.NCERT Fig. 9.6 · p. 156

Line L1 (solid, slope 1/2) crosses the x-axis at (−4, 0) and line L2 (dashed, slope 3) crosses it at (1, 0); they meet at (2, 3). Step 1: arcs mark the inclinations alpha 1 and alpha 2 at the two crossing points on the x-axis. Step 2: an arc at the meeting point marks theta, the angle of the triangle there, with theta = alpha 2 − alpha 1 and tan theta = 1. Step 3: a second arc marks the neighbouring angle phi; theta is 45 degrees and phi is 135 degrees.

Example 9.2NCERT Example 2 · p. 157

Finding a slope from an angle

Two lines meet at $45^{\circ}$ , and one of them has slope 2. What slopes are possible for the other?

Try it yourself first. Then reveal the solution one step at a time.

Why two answers? Through any point there are two lines that make $45^{\circ}$ with a given line, one on each side of it (Fig. 9.7). Check: $∣ \frac{- 3 - 2}{1 - 6} ∣ = 1$ and $∣ \frac{1 / 3 - 2}{1 + 2 / 3} ∣ = 1$ .

3/3Turning forward by $45^{\circ}$ gives slope $- 3$ .

Fig. 9.7One line, one angle, two possible partners.NCERT Fig. 9.7 · p. 158

Axes with a solid blue line of slope 2 through the origin. Step 1 shows only this line and asks which slope makes a 45 degree angle with it. Step 2: a dashed copy turns 45 degrees clockwise to slope 1/3, with an arc marking 45 degrees between the lines. Step 3: a dotted copy turns 45 degrees anticlockwise to slope −3, with a second 45 degree arc, so both answers are visible together.

Example 9.3NCERT Example 3 · p. 158

Using perpendicular slopes

The points $(1, 2)$ and $(5, 4)$ determine one line, and $(3, - 1)$ and $(k, 7)$ another. For which $k$ do the lines meet at right angles?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

The acute angle between

y = 3 x + 1

and

y = - 2 x + 5

30^{\circ}

60^{\circ}

45^{\circ}

135^{\circ}

Show answer

Answer (C) $45^{\circ}$

\tan θ = ∣ \frac{- 2 - 3}{1 + (3) (- 2)} ∣ = ∣ \frac{- 5}{- 5} ∣ = 1

, so

θ = 45^{\circ}

135^{\circ}

is the other angle between the lines, but it is obtuse.

Practice · 11 questions

Practice 9.1: questions with answers

Short answerMediumNCERT Exercise 9.1, Q1
Mark $A (- 3, 4)$ , $B (2, 6)$ , $C (6, - 2)$ and $D (- 2, - 4)$ on a grid, join them in order, and work out the area of $A B C D$ .
Show solution
Answer57 square units
Plotted in order, $A B C D$ is a convex quadrilateral, so the diagonal $A C$ splits it into triangles $A B C$ and $A C D$ .
Triangle $A B C$ by (9.3): the three terms are $- 3 (8) = - 24$ , $2 (- 6) = - 12$ and $6 (- 2) = - 12$ . Their sum is $- 48$ , so the area is $\frac{1}{2} \cdot 48 = 24$ .
Triangle $A C D$ : the terms are $- 3 (2) = - 6$ , $6 (- 8) = - 48$ and $- 2 (6) = - 12$ . Their sum is $- 66$ , so the area is $\frac{1}{2} \cdot 66 = 33$ .
Area of $A B C D$ $= 24 + 33 = 57$ square units.
Short answerMediumNCERT Exercise 9.1, Q2
An equilateral triangle with sides $2 b$ sits with one side on the $x$ -axis, centred at the origin. Where are its corners?
Show solution
The base runs from $(- b, 0)$ to $(b, 0)$ . The third vertex is equidistant from both ends, so it lies on the $y$ -axis at $(0, h)$ .
Its distance from $(b, 0)$ is $2 b$ : $b^{2} + h^{2} = 4 b^{2}$ , so $h = \pm \sqrt{3} b$ .
Vertices: $(- b, 0)$ , $(b, 0)$ and $(0, \sqrt{3} b)$ , or $(0, - \sqrt{3} b)$ for the triangle below the axis.
3
Short answerEasyNCERT Exercise 9.1, Q3
Find the distance between two points that share a coordinate.
1. $A (5, - 3)$ and $B (5, 9)$ , and in general $P (p, y_{1})$ and $Q (p, y_{2})$ (so $P Q$ is parallel to the $y$ -axis).
  Show solution
  The $x$ -difference is 0, so $A B = ∣ 9 - (- 3) ∣ = 12$ . In general $P Q = ∣ y_{2} - y_{1} ∣$ .
2. $C (- 4, 7)$ and $D (6, 7)$ , and in general $P (x_{1}, q)$ and $Q (x_{2}, q)$ (so $P Q$ is parallel to the $x$ -axis).
  Show solution
  The $y$ -difference is 0, so $C D = ∣ 6 - (- 4) ∣ = 10$ . In general $P Q = ∣ x_{2} - x_{1} ∣$ .
Short answerMediumNCERT Exercise 9.1, Q4
Which point of the $y$ -axis is as far from $(5, 2)$ as from $(- 1, 6)$ ?
Show solution
Answer $(0, 1)$
Let the point be $(0, k)$ . Then $25 + (k - 2)^{2} = 1 + (k - 6)^{2}$ .
Expanding: $29 - 4 k + k^{2} = 37 - 12 k + k^{2}$ , so $8 k = 8$ and $k = 1$ .
The point is $(0, 1)$ . Check: both squared distances are $26$ .
Quick answerEasyNCERT Exercise 9.1, Q5
$M$ is the midpoint of $A (- 2, 3)$ and $B (6, - 7)$ . What is the slope of the line joining $(1, 5)$ to $M$ ?
Show answer
Answer -7
The midpoint is $(\frac{- 2 + 6}{2}, \frac{3 - 7}{2}) = (2, - 2)$ . Slope $= \frac{- 2 - 5}{2 - 1} = - 7$ .
ProofMediumNCERT Exercise 9.1, Q6
Use slopes, not Pythagoras' theorem, to show that triangle $A B C$ with $A (1, 2)$ , $B (4, 3)$ , $C (2, 9)$ has a right angle, and say where it is.
Show solution
Slopes: $A B = \frac{3 - 2}{4 - 1} = \frac{1}{3}$ , $B C = \frac{9 - 3}{2 - 4} = - 3$ and $A C = \frac{9 - 2}{2 - 1} = 7$ .
Since $\frac{1}{3} \times (- 3) = - 1$ , $A B ⊥ B C$ . The triangle has a right angle at $B$ .
Short answerMediumNCERT Exercise 9.1, Q7
Start facing along the positive $y$ -axis and turn $60^{\circ}$ anticlockwise. What is the slope of a line pointing in that direction?
Show solution
Answer $- \frac{1}{\sqrt{3}}$
The positive $y$ -axis points at $90^{\circ}$ from the positive $x$ -axis. Turning a further $60^{\circ}$ anticlockwise gives an inclination of $150^{\circ}$ .
So $m = \tan 150^{\circ} = - \tan 30^{\circ} = - \frac{1}{\sqrt{3}}$ .
ProofMediumNCERT Exercise 9.1, Q8
Using slopes only, show that $P Q R S$ with $P (- 1, - 2)$ , $Q (5, 1)$ , $R (6, 5)$ and $S (0, 2)$ is a parallelogram.
Show solution
Slopes: $P Q = \frac{3}{6} = \frac{1}{2}$ , $Q R = \frac{4}{1} = 4$ , $R S = \frac{- 3}{- 6} = \frac{1}{2}$ , $S P = \frac{- 4}{- 1} = 4$ .
So $P Q ∥ R S$ and $Q R ∥ S P$ . The slopes $\frac{1}{2}$ and $4$ differ, so the four points do not lie on one line, and $P Q R S$ is a parallelogram.
Short answerEasyNCERT Exercise 9.1, Q9
What is the inclination of the line through $(2, - \sqrt{3})$ and $(3, 0)$ ?
Show solution
Answer $60^{\circ}$
$m = \frac{0 - (- \sqrt{3})}{3 - 2} = \sqrt{3}$ . So $\tan θ = \sqrt{3}$ with $0^{\circ} \leq θ < 180^{\circ}$ , giving $θ = 60^{\circ}$ .
Long answerHardNCERT Exercise 9.1, Q10
Two lines have slopes $m$ and $3 m$ , and the acute angle $θ$ between them has $\tan θ = \frac{1}{2}$ . Find every possible pair of slopes.
Show solution
Let the slopes be $m$ and $3 m$ . By (9.6), $\frac{3 m - m}{1 + 3 m^{2}} = \pm \frac{1}{2}$ , that is, $4 m = \pm (1 + 3 m^{2})$ .
With $+$ : $3 m^{2} - 4 m + 1 = 0$ , so $m = 1$ or $m = \frac{1}{3}$ .
With $-$ : $3 m^{2} + 4 m + 1 = 0$ , so $m = - 1$ or $m = - \frac{1}{3}$ .
Pairs: $1$ and $3$ ; $\frac{1}{3}$ and $1$ ; $- 1$ and $- 3$ ; $- \frac{1}{3}$ and $- 1$ .
ProofEasyNCERT Exercise 9.1, Q11
Points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ lie on a line of slope $m$ . Show that $y_{2} = y_{1} + m (x_{2} - x_{1})$ . Use this to find the $y$ -coordinate of the point with $x = 7$ on the line through $(1, 3)$ with slope $\frac{2}{3}$ .
Show solution
Answer7
By (9.5), $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ , so $y_{2} - y_{1} = m (x_{2} - x_{1})$ , which rearranges to the result. (If $x_{2} = x_{1}$ the points coincide, since the line is not vertical.)
Here $y = 3 + \frac{2}{3} (7 - 1) = 3 + 4 = 7$ .

9.3 Various Forms of the Equation of a Line

How would you check whether $(2, 7)$ lies on the line through $(0, 1)$ with slope 3? You need a rule that the coordinates of points on the line obey and every other point breaks. That rule, written in $x$ and $y$ , is the equation of the line: a point lies on the line exactly when its coordinates satisfy it. Here the rule turns out to be $y = 3 x + 1$ , and $7 = 3 \cdot 2 + 1$ , so the answer is yes.

What we are told about a line decides which form of its equation is quickest to write: a point and the slope, two points, the slope and an intercept, or both intercepts.

9.3.1 Horizontal and vertical lines

Look at the points $(- 3, 3)$ , $(0, 3)$ and $(2, 3)$ . All that they share is $y = 3$ , and that is the equation of the horizontal line through them (Fig. 9.8). In general, the horizontal line at height $k$ is $y = k$ , with $k > 0$ above the $x$ -axis and $k < 0$ below it; so a line $a$ units below the axis is $y = - a$ . Likewise the vertical line meeting the $x$ -axis at $h$ is $x = h$ , and a line $b$ units left of the $y$ -axis is $x = - b$ . The axes themselves are $y = 0$ and $x = 0$ .

3/3Vertical lines fix $x$ : $x = 4$ and $x = - 3$ .

Fig. 9.8A horizontal line fixes

y

; a vertical line fixes

x

.NCERT Fig. 9.8 · p. 159

A grid from −5 to 5 across and −4 to 4 up. Step 1: a horizontal line 3 units above the x-axis, with three marked points at x = −3, 0 and 2 all at height 3 and a double arrow showing the gap of 3; its equation is y = 3. Step 2: a dashed horizontal line 2 units below the x-axis, labelled y = −2. Step 3: two vertical lines are added, x = 4 to the right of the y-axis and x = −3 to its left.

Example 9.4NCERT Example 4 · p. 160

Lines parallel to the axes

Through $(4, - 1)$ draw one horizontal and one vertical line. Write their equations.

Try it yourself first. Then reveal the solution one step at a time.

9.3.2 Point-slope form

Suppose we know one point $P_{0} (x_{0}, y_{0})$ of a line $L$ and its slope $m$ . Any other point $P (x, y)$ belongs to $L$ if and only if the segment $P_{0} P$ has slope $m$ , that is, $\frac{y - y_{0}}{x - x_{0}} = m$ . Multiplying out removes the fraction, and the result is satisfied by $P_{0}$ too:

point-slope form

y - y_{0} = m (x - x_{0})

(9.7)

Every point of $L$ satisfies (9.7). A point off $L$ does not, because the slope from $P_{0}$ to it is different from $m$ (or, for a point straight above or below $P_{0}$ , the left side is non-zero while the right side is zero). Fig. 9.9 shows why the slope is needed: one point lies on infinitely many lines, and the slope picks out one of them.

2/3For any $P (x, y)$ on the line, $\frac{y - y_{0}}{x - x_{0}} = m$ .

Fig. 9.9A point and a slope, or two points, fix a line.NCERT Fig. 9.10 · p. 160

Step 1: five thin lines of different slopes pass through the point P0(1, 2), showing that one point does not fix a line. Step 2: only the blue line of slope 1/2 remains; a general point P(x, y) = (4, 3.5) on it is joined to P0 by a dashed run x − x0 and rise y − y0, whose ratio is the slope m. Step 3: the fixed point is renamed P1 and a second point P2(−3, 0) is marked on the line; the slope from the two points is (2 − 0)/(1 + 3) = 1/2.

Example 9.5NCERT Example 5 · p. 161

A point and a slope

A line has slope 5 and contains $(3, - 2)$ . Write its equation in general form.

Try it yourself first. Then reveal the solution one step at a time.

9.3.3 Two-point form

Two points hand us the slope, so the two-point form is the point-slope form in disguise. If $L$ passes through $P_{1} (x_{1}, y_{1})$ and $P_{2} (x_{2}, y_{2})$ with $x_{1} \neq x_{2}$ , its slope is $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ by (9.5). (Equivalently, $P (x, y)$ is on $L$ exactly when $P_{1}$ , $P_{2}$ and $P$ are collinear.) Using this slope at $P_{1}$ in (9.7) gives

two-point form

\begin{aligned} y - y_{1} & = m (x - x_{1}), \\ m & = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \end{aligned}

(9.8)

Example 9.6NCERT Example 6 · p. 161

A line through two points

Which line joins $(- 1, 4)$ and $(3, - 4)$ ?

Try it yourself first. Then reveal the solution one step at a time.

9.3.4 Slope-intercept form

The place where a line crosses the $y$ -axis is especially handy: it is the value of $y$ when $x = 0$ . If the crossing point is $(0, c)$ , the number $c$ is the $y$ -intercept of the line; if the line meets the $x$ -axis at $(d, 0)$ , then $d$ is its $x$ -intercept. Intercepts are coordinates, not lengths, so they are negative for crossings below the origin or to its left.

Slope $m$ , $y$ -intercept $c$ . Use (9.7) at the point $(0, c)$ : $y - c = m (x - 0)$ , which tidies up to

slope-intercept form

y = m x + c

(9.9)

Slope $m$ , $x$ -intercept $d$ . The same step at $(d, 0)$ gives

y = m (x - d)

(9.10)

In $y = m x + c$ the slope and the $y$ -intercept can be read straight off the equation, which makes it the handiest form for sketching a line.

Example 9.7NCERT Example 7 · p. 162

Slope with an intercept

A line has slope $- \frac{2}{3}$ . Find its equation if (a) its $y$ -intercept is 4; (b) its $x$ -intercept is $- 3$ .

Try it yourself first. Then reveal the solution one step at a time.

9.3.5 Intercept form

Suppose a line cuts the $x$ -axis at $(a, 0)$ and the $y$ -axis at $(0, b)$ , with $a$ and $b$ both non-zero. Its slope is $\frac{b - 0}{0 - a} = - \frac{b}{a}$ and its $y$ -intercept is $b$ , so $y = - \frac{b}{a} x + b$ . Dividing by $b$ and moving the $x$ -term across gives

intercept form

\frac{x}{a} + \frac{y}{b} = 1

(9.11)

This form cannot describe a line through the origin (both intercepts are 0) or a line parallel to an axis (one intercept is missing). Fig. 9.10 compares the intercept forms on one line, $3 x + 4 y = 12$ .

3/4Intercepts $a = 4$ , $b = 3$ : $\frac{x}{4} + \frac{y}{3} = 1$ .

Fig. 9.10Intercepts are coordinates, so they can be negative.NCERT Fig. 9.13 · p. 163

Step 1: the blue line 3x + 4y = 12 with its y-axis crossing (0, c) = (0, 3) marked; with slope −3/4 the equation is y = mx + c. Step 2: the x-axis crossing (d, 0) = (4, 0) is marked and the equation becomes y = m(x − d). Step 3: both crossings are marked with double arrows a = 4 along the x-axis and b = 3 along the y-axis, giving x/4 + y/3 = 1. Step 4: the blue line is replaced by a dashed line through (−2, 0) and (0, 3), whose intercepts a = −2 and b = 3 give x/(−2) + y/3 = 1.

Example 9.8NCERT Example 8 · p. 163

Two intercepts

A line cuts the $x$ -axis at 4 and the $y$ -axis at $- 6$ . Find its equation.

Try it yourself first. Then reveal the solution one step at a time.

General equation of a line

Every form above rearranges to $A x + B y + C = 0$ , where the numbers $A$ , $B$ are not both zero, and every such equation describes a line. This is the general linear equation of a line.

general equation

A x + B y + C = 0

(9.12)

where

$A, B$: real numbers, not both zero

If $B \neq 0$ , it rearranges to $y = - \frac{A}{B} x - \frac{C}{B}$ , so the slope is $- \frac{A}{B}$ and the $y$ -intercept is $- \frac{C}{B}$ . If $B = 0$ , it is the vertical line $x = - \frac{C}{A}$ . When $A \neq 0$ , the $x$ -intercept is $- \frac{C}{A}$ . For $3 x - 2 y - 12 = 0$ this gives slope $\frac{3}{2}$ , $y$ -intercept $- 6$ and $x$ -intercept 4, matching Example 9.8.

InteractiveLine lab

A grid from −6 to 6 on both axes with two draggable points P and Q that snap to half units. In Equation mode the panel shows the slope as a fraction (or "not defined" for a vertical line), the inclination in degrees, and the equation in point-slope, slope-intercept, intercept and general form, saying which forms do not exist for the current line. In Distance mode a third point R appears with its perpendicular to the line, the foot of the perpendicular, and the distance worked out with the formula. In Two lines mode a second line through R with an adjustable slope appears, and the panel gives the acute angle between the lines and says when they are parallel or perpendicular.

Fig. 9.11Drag two points to make a line and read its slope, inclination and every form of its equation. Then add a third point to measure its distance from the line, or a second line to measure the angle between them.

Quick check

1 of 4

Which line has slope

- 2

and

y

-intercept 3?

y = 3 x - 2

2 x - y + 3 = 0

x + 2 y = 3

2 x + y - 3 = 0

Show answer

Answer (D) $2 x + y - 3 = 0$

2 x + y - 3 = 0

y = - 2 x + 3

. The option

2 x - y + 3 = 0

y = 2 x + 3

, with slope

+ 2

, and

y = 3 x - 2

swaps the two numbers.

Practice · 19 questions

Practice 9.2: questions with answers

1
Quick answerEasyNCERT Exercise 9.2, Q1
Match each line with its equation.
Tap an item on the left, then its match on the right.
Show answer
Answer
the $x$ -axis : $y = 0$
the $y$ -axis : $x = 0$
through $(0, 4)$ , parallel to the $x$ -axis : $y = 4$
through $(- 2, 0)$ , parallel to the $y$ -axis : $x = - 2$
Points on the $x$ -axis have $y = 0$ and points on the $y$ -axis have $x = 0$ . A horizontal line keeps the $y$ -coordinate of its points; a vertical line keeps the $x$ -coordinate.
Short answerEasyNCERT Exercise 9.2, Q2
A line of slope $- \frac{3}{4}$ contains the point $(5, - 2)$ . Write its equation in general form.
Show solution
Answer $3 x + 4 y - 7 = 0$
$y + 2 = - \frac{3}{4} (x - 5)$ . Multiply by 4: $4 y + 8 = - 3 x + 15$ .
So $3 x + 4 y - 7 = 0$ .
Short answerEasyNCERT Exercise 9.2, Q3
Write the line through the origin with inclination $θ$ . What happens when $θ = 90^{\circ}$ ?
Show solution
For $θ \neq 90^{\circ}$ : $y - 0 = \tan θ (x - 0)$ , so $y = x \tan θ$ .
For $θ = 90^{\circ}$ the line is the $y$ -axis, $x = 0$ .
Short answerMediumNCERT Exercise 9.2, Q4
A line through $(- 1, \sqrt{3})$ has inclination $105^{\circ}$ . Find its equation.
Show solution
$\tan 105^{\circ} = \frac{\tan 60^{\circ} + \tan 45^{\circ}}{1 - \tan 60^{\circ} \tan 45^{\circ}} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = - (2 + \sqrt{3})$ .
So $y - \sqrt{3} = - (2 + \sqrt{3}) (x + 1)$ , which gives $y = - (2 + \sqrt{3}) x - 2$ .
Equation: $(2 + \sqrt{3}) x + y + 2 = 0$ .
Short answerEasyNCERT Exercise 9.2, Q5
A line of slope $\frac{3}{2}$ cuts the $x$ -axis 4 units to the right of $O$ . Find it.
Show solution
By (9.10), $y = \frac{3}{2} (x - 4)$ , so $2 y = 3 x - 12$ .
Equation: $3 x - 2 y - 12 = 0$ .
Short answerEasyNCERT Exercise 9.2, Q6
A line with inclination $60^{\circ}$ meets the $y$ -axis at $(0, - 3)$ . Write its equation.
Show solution
$m = \tan 60^{\circ} = \sqrt{3}$ and $c = - 3$ , so $y = \sqrt{3} x - 3$ , that is, $\sqrt{3} x - y - 3 = 0$ .
Short answerEasyNCERT Exercise 9.2, Q7
Write the equation of the line joining $(- 3, - 2)$ to $(1, 6)$ .
Show solution
$m = \frac{6 + 2}{1 + 3} = 2$ , so $y + 2 = 2 (x + 3)$ .
Equation: $2 x - y + 4 = 0$ . Check: $2 (1) - 6 + 4 = 0$ .
Short answerMediumNCERT Exercise 9.2, Q8
In triangle $A B C$ , $A = (1, 4)$ , $B = (- 3, 2)$ and $C = (5, - 2)$ . The median from $B$ runs to the midpoint of $A C$ . Find its equation.
Show solution
The midpoint of $A C$ is $(3, 1)$ . The median joins $B (- 3, 2)$ to $(3, 1)$ , with slope $\frac{1 - 2}{3 + 3} = - \frac{1}{6}$ .
$y - 2 = - \frac{1}{6} (x + 3)$ , so $6 y - 12 = - x - 3$ .
Equation: $x + 6 y - 9 = 0$ .
Short answerMediumNCERT Exercise 9.2, Q9
A line passes through $(2, - 4)$ and is at right angles to the segment from $(- 1, 3)$ to $(4, 5)$ . Write its equation.
Show solution
The given slope is $\frac{5 - 3}{4 + 1} = \frac{2}{5}$ , so the required slope is $- \frac{5}{2}$ .
$y + 4 = - \frac{5}{2} (x - 2)$ , so $2 y + 8 = - 5 x + 10$ .
Equation: $5 x + 2 y - 2 = 0$ .
Long answerHardNCERT Exercise 9.2, Q10
The segment from $A (- 1, 2)$ to $B (3, 4)$ is cut in the ratio $m : n$ (counted from $A$ ) by a line that crosses it at right angles. Find that line.
Show solution
The dividing point is $(\frac{3 m - n}{m + n}, \frac{4 m + 2 n}{m + n})$ . The slope of $A B$ is $\frac{2}{4} = \frac{1}{2}$ , so the required slope is $- 2$ .
$y - \frac{4 m + 2 n}{m + n} = - 2 (x - \frac{3 m - n}{m + n})$ . Multiply by $m + n$ and simplify:
$2 (m + n) x + (m + n) y = 10 m$ . (Check with $m = n$ : the line passes through the midpoint $(1, 3)$ .)
Short answerMediumNCERT Exercise 9.2, Q11
A line through $(- 4, 7)$ makes the same non-zero intercept on both axes. Find it.
Show solution
The line is $x + y = a$ . It passes through $(- 4, 7)$ , so $a = 3$ .
Equation: $x + y = 3$ . (If both intercepts were allowed to be 0, the line $7 x + 4 y = 0$ through the origin would also qualify.)
Long answerMediumNCERT Exercise 9.2, Q12
The two intercepts of a line through $(1, 4)$ add up to 10. Find all such lines.
Show solution
The line is $\frac{x}{a} + \frac{y}{10 - a} = 1$ . Substituting $(1, 4)$ : $\frac{1}{a} + \frac{4}{10 - a} = 1$ .
So $10 - a + 4 a = a (10 - a)$ , which gives $a^{2} - 7 a + 10 = 0$ and $a = 2$ or $a = 5$ .
$a = 2$ , $b = 8$ : $4 x + y = 8$ . $a = 5$ , $b = 5$ : $x + y = 5$ .
Short answerMediumNCERT Exercise 9.2, Q13
Line $ℓ$ has inclination $\frac{π}{6}$ and $y$ -intercept $- 1$ . Write its equation, and the equation of the line parallel to $ℓ$ with $y$ -intercept 3.
Show solution
$m = \frac{1}{\sqrt{3}}$ and $c = - 1$ : $y = \frac{x}{\sqrt{3}} - 1$ , that is, $x - \sqrt{3} y - \sqrt{3} = 0$ .
Parallel line with $c = 3$ : $y = \frac{x}{\sqrt{3}} + 3$ , that is, $x - \sqrt{3} y + 3 \sqrt{3} = 0$ .
Short answerMediumNCERT Exercise 9.2, Q14
The point of a certain line closest to the origin is $(3, - 4)$ . Write down the line's equation.
Show solution
The closest point is the foot of the perpendicular from $O$ . The slope of $O$ to $(3, - 4)$ is $- \frac{4}{3}$ , so the line has slope $\frac{3}{4}$ and passes through $(3, - 4)$ .
$y + 4 = \frac{3}{4} (x - 3)$ , so $4 y + 16 = 3 x - 9$ .
Equation: $3 x - 4 y - 25 = 0$ .
NumericalMediumNCERT Exercise 9.2, Q15
A steel rail is 12.000 m long at 10 °C and 12.0036 m long at 35 °C. Assuming its length $L$ (in metres) is a linear function of the temperature $T$ (in °C), express $L$ in terms of $T$ and find the length at 45 °C.
Show solution
Slope: $\frac{12.0036 - 12.000}{35 - 10} = 0.000144$ metre per degree.
$L - 12 = 0.000144 (T - 10)$ , so $L = 0.000144 T + 11.99856$ .
At 45 °C: $L = 12 + 0.000144 \times 35 = 12.00504$ m. This is why rails are laid with small gaps.
Quick answerMediumNCERT Exercise 9.2, Q16
A juice stall sells 450 glasses a week at ₹30 a glass and 390 glasses a week at ₹35. Assuming sales fall linearly as the price rises, how many glasses would it sell at ₹40?
glasses
Show answer
Answer 330 glasses
Treat (price, sales) as points: $(30, 450)$ and $(35, 390)$ . Slope $= \frac{390 - 450}{5} = - 12$ glasses per rupee. At ₹40: $390 - 12 \times 5 = 330$ .
17
ProofMediumNCERT Exercise 9.2, Q17
A segment of a line lies between the two axes.
1. Its midpoint is $(3, - 2)$ . Find the equation of the line.
  Show solution
  The ends are $(a, 0)$ and $(0, b)$ with midpoint $(\frac{a}{2}, \frac{b}{2}) = (3, - 2)$ , so $a = 6$ and $b = - 4$ .
  $\frac{x}{6} + \frac{y}{- 4} = 1$ , that is, $2 x - 3 y = 12$ .
2. Show that if the midpoint is $(p, q)$ , with $p, q \neq 0$ , the line is $\frac{x}{p} + \frac{y}{q} = 2$ .
  Show solution
  As in (a), the intercepts are $2 p$ and $2 q$ , so $\frac{x}{2 p} + \frac{y}{2 q} = 1$ . Multiply by 2.
Long answerHardNCERT Exercise 9.2, Q18
A line meets the $x$ -axis at $X$ and the $y$ -axis at $Y$ . The point $R (h, k)$ on $X Y$ has $X R : R Y = 2 : 3$ . Find the line.
Show solution
By (9.2), the dividing point is $(\frac{2 \cdot 0 + 3 p}{5}, \frac{2 q + 3 \cdot 0}{5}) = (\frac{3 p}{5}, \frac{2 q}{5})$ .
So $p = \frac{5 h}{3}$ and $q = \frac{5 k}{2}$ , and the line is $\frac{3 x}{5 h} + \frac{2 y}{5 k} = 1$ .
Equation: $3 k x + 2 h y = 5 h k$ .
ProofEasyNCERT Exercise 9.2, Q19
Find the line through $(- 1, - 4)$ and $(2, 2)$ , and use it to decide whether $(5, 8)$ lies on the same line.
Show solution
Through $(- 1, - 4)$ and $(2, 2)$ : slope 2, so $y + 4 = 2 (x + 1)$ , that is, $y = 2 x - 2$ .
For $(5, 8)$ : $2 (5) - 2 = 8$ . The third point satisfies the equation, so all three lie on one line.

9.4 Distance of a Point From a Line

How far is a village from a straight highway? There are many paths from the village to the road, but the shortest one meets the road at a right angle; any slanting path is the hypotenuse of a right triangle and so is longer. That is why the distance of a point from a line means the length of the perpendicular dropped from the point to the line (Fig. 9.12).

3/4Area of $P Q R$ is 5 and $Q R = 2 \sqrt{5}$ .

Fig. 9.12The perpendicular is the height of triangle

P Q R

on the base

Q R

.NCERT Fig. 9.14 · p. 164

The line x − 2y + 4 = 0 crosses the x-axis at Q(−4, 0) and the y-axis at R(0, 2). Step 1: from P(3, 1) three dashed slanting paths run to the line, with the solid perpendicular PM the shortest. Step 2: PM is marked with a right angle at the foot M(2, 3), and d = PM. Step 3: triangle PQR is shaded; its area is 5 and the base QR is 2√5, with PM as the height on that base. Step 4: d = 2 × 5 ÷ 2√5 = √5.

DerivationDistance of

(x_{1}, y_{1})

from

A x + B y + C = 0

Let $P (x_{1}, y_{1})$ be the point and $L : A x + B y + C = 0$ the line, with $A$ , $B$ and $C$ all non-zero, so that $L$ meets the axes at two different points $Q (- \frac{C}{A}, 0)$ and $R (0, - \frac{C}{B})$ . Let $M$ be the foot of the perpendicular from $P$ , so $d = P M$ .

Take $Q R$ as the base of triangle $P Q R$ . Its height is then $P M$ , so

d = \frac{2 \cdot area (P Q R)}{Q R}

Write $S = A x_{1} + B y_{1} + C$ and $k = \frac{C}{A B}$ , so that $Q = (- k B, 0)$ and $R = (0, - k A)$ . Putting the three vertices into (9.3), the bracket is $x_{1} (k A) - k B (- k A - y_{1}) = k (A x_{1} + B y_{1} + k A B)$ , which is $k S$ because $k A B = C$ . So

2 \cdot area = ∣ k ∣ ∣ S ∣

By the distance formula, $Q R^{2} = k^{2} B^{2} + k^{2} A^{2}$ , so $Q R = ∣ k ∣ \sqrt{A^{2} + B^{2}}$ .

Dividing, the factor $∣ k ∣$ cancels and we are left with (9.13). The formula also holds when $L$ passes through the origin or is parallel to an axis; for instance the distance of $(x_{1}, y_{1})$ from $x - k = 0$ comes out as $∣ x_{1} - k ∣$ , as it should.

distance of a point from a line

d = \frac{∣ A x_{1} + B y_{1} + C ∣}{\sqrt{A^{2} + B^{2}}}

(9.13)

For the figure, $P (3, 1)$ and $x - 2 y + 4 = 0$ give $d = \frac{∣ 3 - 2 + 4 ∣}{\sqrt{1 + 4}} = \frac{5}{\sqrt{5}} = \sqrt{5}$ . In particular, the distance of the origin from $A x + B y + C = 0$ is $\frac{∣ C ∣}{\sqrt{A^{2} + B^{2}}}$ .

9.4.1 Distance between two parallel lines

Parallel lines have the same slope, so we can write them as $y = m x + c_{1}$ and $y = m x + c_{2}$ . The gap between them is the same everywhere, so we may measure it from any convenient point of the first line, such as its $y$ -axis point $A (0, c_{1})$ (Fig. 9.13). Write the second line as $m x - y + c_{2} = 0$ and apply (9.13) at $A$ :

d = \frac{∣ m \cdot 0 - c_{1} + c_{2} ∣}{\sqrt{m^{2} + 1}}

parallel lines in slope form

d = \frac{∣ c_{1} - c_{2} ∣}{\sqrt{1 + m^{2}}}

(9.14)

3/4 $d = \frac{∣ 3 - (- 2) ∣}{\sqrt{1 + 1 / 4}} = 2 \sqrt{5}$ .

Fig. 9.13Measure the gap from any point of one line.NCERT Fig. 9.15 · p. 166

Step 1: the parallel lines y = x/2 + 3 (solid) and y = x/2 − 2 (dashed). Step 2: the point A(0, 3) where the upper line meets the y-axis is marked. Step 3: the perpendicular from A meets the lower line at (2, −1), with a right-angle mark; its length is 2√5. Step 4: a second perpendicular from (4, 5) on the upper line meets the lower line at (6, 1) and has the same length, about 4.47.

If both lines are written in general form with the same $A$ and $B$ , so that they differ only in their constants $C_{1}$ and $C_{2}$ , the same idea (try it) gives

parallel lines in general form

d = \frac{∣ C_{1} - C_{2} ∣}{\sqrt{A^{2} + B^{2}}}

(9.15)

Example 9.9NCERT Example 9 · p. 167

Distance of a point from a line

How far is the point $(- 2, 3)$ from the line $5 x - 12 y + 7 = 0$ ?

Try it yourself first. Then reveal the solution one step at a time.

Example 9.10NCERT Example 10 · p. 167

Gap between parallel lines

How far apart are the parallel lines $6 x + 8 y - 9 = 0$ and $6 x + 8 y + 11 = 0$ ?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Find the distance of

(1, - 2)

from the line

3 x + 4 y - 10 = 0

Show answer

Answer 3

d = \frac{∣ 3 (1) + 4 (- 2) - 10 ∣}{\sqrt{9 + 16}} = \frac{∣ - 15 ∣}{5} = 3

. Without the modulus you would get

- 3

, but a distance is never negative.

Practice · 17 questions

Practice 9.3: questions with answers

1
Short answerEasyNCERT Exercise 9.3, Q1
Rewrite each equation as $y = m x + c$ where possible, and state $m$ and $c$ .
1. $2 x - 5 y = 0$
  Show solution
  $y = \frac{2}{5} x$ : slope $\frac{2}{5}$ , $y$ -intercept 0.
2. $4 x + 2 y - 7 = 0$
  Show solution
  $y = - 2 x + \frac{7}{2}$ : slope $- 2$ , $y$ -intercept $\frac{7}{2}$ .
3. $y - 3 = 0$
  Show solution
  $y = 0 \cdot x + 3$ : slope 0, $y$ -intercept 3.
4. $x = 4$
  Show solution
  Not possible: the line is vertical, so it has no slope and does not meet the $y$ -axis.
2
Short answerEasyNCERT Exercise 9.3, Q2
Rewrite each equation as $\frac{x}{a} + \frac{y}{b} = 1$ where possible, and state the intercepts.
1. $5 x + 3 y - 15 = 0$
  Show solution
  $5 x + 3 y = 15$ , so $\frac{x}{3} + \frac{y}{5} = 1$ : intercepts 3 and 5.
2. $2 x - 7 y = 14$
  Show solution
  $\frac{x}{7} + \frac{y}{- 2} = 1$ : intercepts 7 and $- 2$ .
3. $2 x + 5 = 0$
  Show solution
  $x = - \frac{5}{2}$ , a vertical line: $x$ -intercept $- \frac{5}{2}$ and no $y$ -intercept, so there is no intercept form.
Quick answerEasyNCERT Exercise 9.3, Q3
How far is $(2, - 3)$ from the line $3 (x - 1) = 4 (y + 2)$ ?
Show answer
Answer 1.4
The line is $3 x - 4 y - 11 = 0$ . So $d = \frac{∣ 6 + 12 - 11 ∣}{5} = \frac{7}{5} = 1.4$ .
Short answerMediumNCERT Exercise 9.3, Q4
Which points of the $x$ -axis are 4 units from the line $\frac{x}{6} + \frac{y}{8} = 1$ ?
Show solution
The line is $4 x + 3 y - 24 = 0$ . For $(h, 0)$ : $\frac{∣ 4 h - 24 ∣}{5} = 4$ , so $4 h - 24 = \pm 20$ .
$h = 11$ or $h = 1$ . The points are $(11, 0)$ and $(1, 0)$ .
5
Short answerMediumNCERT Exercise 9.3, Q5
Find the distance between each pair of parallel lines.
1. $12 x - 5 y + 20 = 0$ and $12 x - 5 y - 6 = 0$
  Show solution
  $d = \frac{∣ 20 - (- 6) ∣}{\sqrt{144 + 25}} = \frac{26}{13} = 2$ .
2. $a (x - y) + p = 0$ and $a (x - y) - q = 0$ , with $a \neq 0$
  Show solution
  Here $A = a$ , $B = - a$ , $C_{1} = p$ , $C_{2} = - q$ , so $d = \frac{∣ p + q ∣}{∣ a ∣ \sqrt{2}}$ .
Short answerEasyNCERT Exercise 9.3, Q6
Which line through $(1, - 4)$ never meets $5 x + 2 y - 7 = 0$ ?
Show solution
A parallel line is $5 x + 2 y + k = 0$ . Through $(1, - 4)$ : $5 - 8 + k = 0$ , so $k = 3$ .
Equation: $5 x + 2 y + 3 = 0$ .
Short answerMediumNCERT Exercise 9.3, Q7
A line crosses $2 x + 3 y - 6 = 0$ at right angles and has $y$ -intercept $- 2$ . Find it.
Show solution
The given slope is $- \frac{2}{3}$ , so the required slope is $\frac{3}{2}$ , and $y = \frac{3}{2} x - 2$ .
Equation: $3 x - 2 y - 4 = 0$ .
Short answerMediumNCERT Exercise 9.3, Q8
Find the angles between the lines $x - \sqrt{3} y + 1 = 0$ and $x - y = 3$ .
Show solution
The inclinations are $30^{\circ}$ and $45^{\circ}$ , so the acute angle is $15^{\circ}$ .
By (9.6): $\tan θ = \frac{1 - 1 / \sqrt{3}}{1 + 1 / \sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3} = \tan 15^{\circ}$ , which agrees.
The angles are $15^{\circ}$ and $165^{\circ}$ .
Quick answerMediumNCERT Exercise 9.3, Q9
The line through $(2, k)$ and $(- 1, 5)$ is perpendicular to $3 x + 2 y - 8 = 0$ . Find $k$ .
Show answer
Answer 7
The given slope is $- \frac{3}{2}$ , so the required slope is $\frac{2}{3}$ . Then $\frac{k - 5}{2 - (- 1)} = \frac{2}{3}$ gives $k - 5 = 2$ , so $k = 7$ .
ProofMediumNCERT Exercise 9.3, Q10
Prove that the line through $(x_{1}, y_{1})$ perpendicular to $A x + B y + C = 0$ is $B (x - x_{1}) - A (y - y_{1}) = 0$ .
Show solution
If $A, B \neq 0$ : the given slope is $- \frac{A}{B}$ , so the required slope is $\frac{B}{A}$ . Then $y - y_{1} = \frac{B}{A} (x - x_{1})$ , which rearranges to $B (x - x_{1}) - A (y - y_{1}) = 0$ .
If $B = 0$ the given line is vertical, the required line is $y = y_{1}$ , and the formula reduces to $- A (y - y_{1}) = 0$ . If $A = 0$ the given line is horizontal, the required line is $x = x_{1}$ , and the formula reduces to $B (x - x_{1}) = 0$ .
Long answerMediumNCERT Exercise 9.3, Q11
Through $(1, - 2)$ pass two lines that cross at $60^{\circ}$ . If one has slope $\sqrt{3}$ , what can the other be?
Show solution
The first line has inclination $60^{\circ}$ . The other differs by $60^{\circ}$ , so its inclination is $0^{\circ}$ or $120^{\circ}$ , and its slope is $0$ or $- \sqrt{3}$ .
Slope 0: $y = - 2$ . Slope $- \sqrt{3}$ : $y + 2 = - \sqrt{3} (x - 1)$ , that is, $\sqrt{3} x + y + 2 - \sqrt{3} = 0$ .
Check with (9.6): both give $\tan θ = \sqrt{3}$ .
Short answerMediumNCERT Exercise 9.3, Q12
Every point of a certain line is equidistant from $(- 2, 5)$ and $(4, 1)$ . Find the line.
Show solution
The points equidistant from two given points form the perpendicular bisector of the segment joining them. Midpoint $(1, 3)$ . The segment's slope is $\frac{1 - 5}{4 + 2} = - \frac{2}{3}$ , so the bisector's slope is $\frac{3}{2}$ .
$y - 3 = \frac{3}{2} (x - 1)$ , so $3 x - 2 y + 3 = 0$ .
Short answerMediumNCERT Exercise 9.3, Q13
Drop a perpendicular from $(4, 1)$ to the line $2 x - y + 3 = 0$ . Where does it land?
Show solution
The perpendicular through $(4, 1)$ has slope $- \frac{1}{2}$ : $x + 2 y - 6 = 0$ .
Solve with $2 x - y + 3 = 0$ : from the first, $x = 6 - 2 y$ , so $12 - 5 y + 3 = 0$ , $y = 3$ , $x = 0$ .
The foot is $(0, 3)$ .
Short answerMediumNCERT Exercise 9.3, Q14
The point of the line $y = m x + c$ nearest to the origin is $(3, - 1)$ . Find $m$ and $c$ .
Show solution
The slope of $O$ to $(3, - 1)$ is $- \frac{1}{3}$ , so $m = 3$ .
The point lies on the line: $- 1 = 3 (3) + c$ , so $c = - 10$ .
ProofHardNCERT Exercise 9.3, Q15
Let $p$ be the distance from the origin to the line through $(a, 0)$ and $(0, a \tan θ)$ , and $q$ the distance from the origin to the line through $(a, 0)$ and $(0, a \cot θ)$ , where $0^{\circ} < θ < 90^{\circ}$ and $a > 0$ . Prove that $p^{2} + q^{2} = a^{2}$ .
Show solution
First line: $\frac{x}{a} + \frac{y}{a \tan θ} = 1$ , that is, $x \tan θ + y - a \tan θ = 0$ . So $p = \frac{a \tan θ}{\sqrt{\tan^{2} θ + 1}} = \frac{a \tan θ}{\sec θ} = a \sin θ$ .
In the same way, the second line is $x \cot θ + y - a \cot θ = 0$ , and $q = \frac{a \cot θ}{\csc θ} = a \cos θ$ .
Hence $p^{2} + q^{2} = a^{2} (\sin^{2} θ + \cos^{2} θ) = a^{2}$ .
Long answerMediumNCERT Exercise 9.3, Q16
For $A (1, 4)$ , $B (- 2, - 1)$ and $C (4, 1)$ , find the line through $A$ perpendicular to $B C$ , and the distance from $A$ to $B C$ .
Show solution
$B C$ has slope $\frac{1 + 1}{4 + 2} = \frac{1}{3}$ : $y + 1 = \frac{1}{3} (x + 2)$ , that is, $x - 3 y - 1 = 0$ .
The altitude has slope $- 3$ : $y - 4 = - 3 (x - 1)$ , that is, $3 x + y - 7 = 0$ .
Length $= \frac{∣ 1 - 12 - 1 ∣}{\sqrt{10}} = \frac{12}{\sqrt{10}} = \frac{6 \sqrt{10}}{5} \approx 3.79$ .
ProofMediumNCERT Exercise 9.3, Q17
Let the intercepts of a line be $a$ and $b$ (both non-zero), and let $p$ be its distance from the origin. Prove that $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$ , then find $p$ when $a = 6$ and $b = 8$ .
Show solution
The line is $b x + a y - a b = 0$ , so $p = \frac{∣ a b ∣}{\sqrt{a^{2} + b^{2}}}$ and $\frac{1}{p^{2}} = \frac{a^{2} + b^{2}}{a^{2} b^{2}} = \frac{1}{b^{2}} + \frac{1}{a^{2}}$ .
For $a = 6$ , $b = 8$ : $p = \frac{48}{10} = 4.8$ .

Miscellaneous Examples

These examples combine the tools of the chapter: intersections, slopes, midpoints and distances. Lines that share one common point are called concurrent. For three lines the quick test is to solve two of them together and substitute that point into the third.

Example 9.11NCERT Example 11 · p. 168

Three concurrent lines

The lines $x + 2 y - 5 = 0$ , $3 x - y - 1 = 0$ and $k x + 3 y - 10 = 0$ pass through one point. Find $k$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 9.12NCERT Example 12 · p. 169

Distance along a given direction

Starting at $P (2, 1)$ , you move along the line of inclination $45^{\circ}$ until you reach the line $2 x + y - 11 = 0$ . How far do you move?

Try it yourself first. Then reveal the solution one step at a time.

Example 9.13NCERT Example 13 · p. 169

Image in a line mirror

The line $x + 2 y - 4 = 0$ acts as a mirror. Where does the reflection of $P (3, - 2)$ appear?

Try it yourself first. Then reveal the solution one step at a time.

3/3Go the same distance beyond $N$ : $Q (5, 2)$ .

Fig. 9.14The mirror is the perpendicular bisector of

P Q

.NCERT Fig. 9.17 · p. 169

The mirror line x + 2y − 4 = 0 runs down from left to right. Step 1 shows only the point P(3, −2) below the mirror. Step 2 adds the dashed perpendicular through P with slope 2, meeting the mirror at N(4, 0) with a right-angle mark. Step 3 continues the dashed line an equal distance beyond N to the image Q(5, 2), with equal tick marks on PN and NQ.

Example 9.14NCERT Example 14 · p. 170

A triangle cut off by two lines and an axis

Lines $y = m_{1} x + c_{1}$ and $y = m_{2} x + c_{2}$ , with $m_{1} \neq m_{2}$ and $m_{1}, m_{2} \neq 0$ , form a triangle with the $x$ -axis. Show that its area is $\frac{(m_{1} c_{2} - m_{2} c_{1})^{2}}{2 ∣ m_{1} m_{2} (m_{1} - m_{2}) ∣}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 9.15NCERT Example 15 · p. 171

A segment bisected at a given point

A line meets $x - y + 1 = 0$ at $A$ and $3 x + y - 15 = 0$ at $B$ , and the midpoint of $A B$ is $(2, 4)$ . Find the equation of the line.

Try it yourself first. Then reveal the solution one step at a time.

Example 9.16NCERT Example 16 · p. 172

Equidistant from two lines

A point moves so that it is always as far from $x + 2 y - 3 = 0$ as from $2 x + y + 1 = 0$ . Show that its path is made of straight lines, and find them.

Try it yourself first. Then reveal the solution one step at a time.

Recap

Straight Lines: key points and summary

Distance $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ , the section and midpoint formulas, and the triangle-area formula; zero area means collinear points (Section 9.1).
The inclination $θ$ is measured anticlockwise from the positive $x$ -axis, $0^{\circ} \leq θ < 180^{\circ}$ , and the slope is $m = \tan θ$ (Section 9.2).
Horizontal lines have slope 0; vertical lines have no slope.
Through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ for $x_{1} \neq x_{2}$ ((9.5)).
For lines that are not vertical: equal slopes mean parallel, and $m_{1} m_{2} = - 1$ means perpendicular. Three points lie on one line exactly when $A B$ and $B C$ have the same slope (Section 9.2.2).
The acute angle between two lines satisfies $\tan θ = ∣ \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} ∣$ ; the other angle is $180^{\circ} - θ$ ((9.6)).
Horizontal lines are $y = \pm a$ and vertical lines are $x = \pm b$ ; the axes are $y = 0$ and $x = 0$ (Section 9.3.1).
Point-slope form $y - y_{0} = m (x - x_{0})$ ((9.7)) and two-point form ((9.8)).
Slope-intercept forms $y = m x + c$ and $y = m (x - d)$ ((9.9)); intercept form $\frac{x}{a} + \frac{y}{b} = 1$ ((9.11)).
Every line is $A x + B y + C = 0$ ( $A$ , $B$ not both 0), with slope $- \frac{A}{B}$ , $x$ -intercept $- \frac{C}{A}$ and $y$ -intercept $- \frac{C}{B}$ ((9.12)).
The distance of $(x_{1}, y_{1})$ from $A x + B y + C = 0$ is $\frac{∣ A x_{1} + B y_{1} + C ∣}{\sqrt{A^{2} + B^{2}}}$ ((9.13)).
Two parallel lines that share $A$ and $B$ and have constants $C_{1}$ , $C_{2}$ are $\frac{∣ C_{1} - C_{2} ∣}{\sqrt{A^{2} + B^{2}}}$ apart; match the coefficients first ((9.15)).

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 9.1, Q1	Practice 9.1 · Q1
NCERT Exercise 9.1, Q2	Practice 9.1 · Q2
NCERT Exercise 9.1, Q3	Practice 9.1 · Q3
NCERT Exercise 9.1, Q4	Practice 9.1 · Q4
NCERT Exercise 9.1, Q5	Practice 9.1 · Q5
NCERT Exercise 9.1, Q6	Practice 9.1 · Q6
NCERT Exercise 9.1, Q7	Practice 9.1 · Q7
NCERT Exercise 9.1, Q8	Practice 9.1 · Q8
NCERT Exercise 9.1, Q9	Practice 9.1 · Q9
NCERT Exercise 9.1, Q10	Practice 9.1 · Q10
NCERT Exercise 9.1, Q11	Practice 9.1 · Q11
NCERT Exercise 9.2, Q1	Practice 9.2 · Q1
NCERT Exercise 9.2, Q2	Practice 9.2 · Q2
NCERT Exercise 9.2, Q3	Practice 9.2 · Q3
NCERT Exercise 9.2, Q4	Practice 9.2 · Q4
NCERT Exercise 9.2, Q5	Practice 9.2 · Q5
NCERT Exercise 9.2, Q6	Practice 9.2 · Q6
NCERT Exercise 9.2, Q7	Practice 9.2 · Q7
NCERT Exercise 9.2, Q8	Practice 9.2 · Q8
NCERT Exercise 9.2, Q9	Practice 9.2 · Q9
NCERT Exercise 9.2, Q10	Practice 9.2 · Q10
NCERT Exercise 9.2, Q11	Practice 9.2 · Q11
NCERT Exercise 9.2, Q12	Practice 9.2 · Q12
NCERT Exercise 9.2, Q13	Practice 9.2 · Q13
NCERT Exercise 9.2, Q14	Practice 9.2 · Q14
NCERT Exercise 9.2, Q15	Practice 9.2 · Q15
NCERT Exercise 9.2, Q16	Practice 9.2 · Q16
NCERT Exercise 9.2, Q17	Practice 9.2 · Q17
NCERT Exercise 9.2, Q18	Practice 9.2 · Q18
NCERT Exercise 9.2, Q19	Practice 9.2 · Q19
NCERT Exercise 9.3, Q1	Practice 9.3 · Q1
NCERT Exercise 9.3, Q2	Practice 9.3 · Q2
NCERT Exercise 9.3, Q3	Practice 9.3 · Q3
NCERT Exercise 9.3, Q4	Practice 9.3 · Q4
NCERT Exercise 9.3, Q5	Practice 9.3 · Q5
NCERT Exercise 9.3, Q6	Practice 9.3 · Q6
NCERT Exercise 9.3, Q7	Practice 9.3 · Q7
NCERT Exercise 9.3, Q8	Practice 9.3 · Q8
NCERT Exercise 9.3, Q9	Practice 9.3 · Q9
NCERT Exercise 9.3, Q10	Practice 9.3 · Q10
NCERT Exercise 9.3, Q11	Practice 9.3 · Q11
NCERT Exercise 9.3, Q12	Practice 9.3 · Q12
NCERT Exercise 9.3, Q13	Practice 9.3 · Q13
NCERT Exercise 9.3, Q14	Practice 9.3 · Q14
NCERT Exercise 9.3, Q15	Practice 9.3 · Q15
NCERT Exercise 9.3, Q16	Practice 9.3 · Q16
NCERT Exercise 9.3, Q17	Practice 9.3 · Q17
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercises · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous exercises · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous exercises · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous exercises · Q10
NCERT Miscellaneous Exercise, Q11	Miscellaneous exercises · Q11
NCERT Miscellaneous Exercise, Q12	Miscellaneous exercises · Q12
NCERT Miscellaneous Exercise, Q13	Miscellaneous exercises · Q13
NCERT Miscellaneous Exercise, Q14	Miscellaneous exercises · Q14
NCERT Miscellaneous Exercise, Q15	Miscellaneous exercises · Q15
NCERT Miscellaneous Exercise, Q16	Miscellaneous exercises · Q16
NCERT Miscellaneous Exercise, Q17	Miscellaneous exercises · Q17
NCERT Miscellaneous Exercise, Q18	Miscellaneous exercises · Q18
NCERT Miscellaneous Exercise, Q19	Miscellaneous exercises · Q19
NCERT Miscellaneous Exercise, Q20	Miscellaneous exercises · Q20
NCERT Miscellaneous Exercise, Q21	Miscellaneous exercises · Q21
NCERT Miscellaneous Exercise, Q22	Miscellaneous exercises · Q22
NCERT Miscellaneous Exercise, Q23	Miscellaneous exercises · Q23

Quick check: MCQs on Straight Lines

Tools from earlier classes3 questions
9.2.2 Conditions for parallel and perpendicular lines4 questions
9.2.3 Angle between two lines3 questions
General equation of a line4 questions
9.4.1 Distance between two parallel lines3 questions
Miscellaneous Examples3 questions

Glossary · 7

Key terms (English–Hindi)

collinearसंरेखsanrekh: Lying on one straight line. Three points are collinear when the triangle they form has zero area.First seen in the chapter →
concurrent linesसंगामी रेखाएँsangami rekhaen: Three or more lines that all pass through one common point.First seen in the chapter →

general linear equationव्यापक रैखिक समीकरणvyapak rekhik samikaran: An equation $A x + B y + C = 0$ , where $A$ , $B$ are not both zero. Every line can be written this way.First seen in the chapter →

inclinationझुकावjhukav: The angle $θ$ from the positive $x$ -axis, turning anticlockwise, to a line; $0^{\circ} \leq θ < 180^{\circ}$ .First seen in the chapter →
interceptअंतःखंडantahkhand: The coordinate where a line crosses an axis: $c$ if it meets the $y$ -axis at $(0, c)$ , $d$ if it meets the $x$ -axis at $(d, 0)$ .First seen in the chapter →

perpendicular distanceलंबवत दूरीlambvat doori: How far a point is from a line, measured along the perpendicular dropped to it, which is the shortest route.First seen in the chapter →

slopeढालdhaal: $m = \tan θ$ for a line of inclination $θ \neq 90^{\circ}$ ; equal to rise over run between two of its points.

Straight Lines: Class 11 Maths · Chapter 9

What you will learn in Straight Lines

9.1 Introduction

Tools from earlier classes

9.2 Slope of a Line

9.2.1 Slope from the coordinates of two points

9.2.2 Conditions for parallel and perpendicular lines

9.2.3 Angle between two lines

Practice 9.1: questions with answers

9.3 Various Forms of the Equation of a Line

9.3.1 Horizontal and vertical lines

9.3.2 Point-slope form

9.3.3 Two-point form

9.3.4 Slope-intercept form

9.3.5 Intercept form

General equation of a line

Practice 9.2: questions with answers

9.4 Distance of a Point From a Line

9.4.1 Distance between two parallel lines

Practice 9.3: questions with answers

Miscellaneous Examples

Practice questions with answers (NCERT “Miscellaneous exercises”)

Straight Lines: key points and summary

Find your NCERT exercise question

Quick check: MCQs on Straight Lines

Key terms (English–Hindi)

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