MHT-CET Maths · Straight Line

Slope, Angle Between Lines and Rotation

Slope is tan of the inclination; tan θ = |(m₁ − m₂)/(1 + m₁m₂)| gives the angle between two lines, and the same formula, solved for m, gives the lines through a point at a given angle, the rotated line, the angle bisector and the reflected line.

All Straight Line notes MHT-CET Maths strategy

Why this matters

19 PYQs at 26% HARD — the chapter's hardest page. The recurring stems are a perpendicular-slope condition, an acute angle between two lines (once between the diagonals of a parallelogram), lines through a point at 45° or 60° to a given line, a line rotated about a point by 15° or 45°, and the bisector of angle PQR set in two consecutive years. All five HARD questions are the angle formula run backwards — solve for the unknown slope — and the wrong answer is always the second root.

Concept 1 of 5: Slope and Inclination: Parallel Means m₁ = m₂, Perpendicular Means m₁m₂ = −1

The slope

m = \tan\theta

is rise over run; two points give it as

\dfrac{y_2 - y_1}{x_2 - x_1}

, and

ax + by + c = 0

gives

-\dfrac{a}{b}

. Parallel lines share a slope; perpendicular slopes multiply to

-1

Definition

$2x - 3y + 17 = 0$ has slope $\dfrac23$ ; the line through $(7, 17)$ and $(15, \beta)$ has slope $\dfrac{\beta - 17}{8}$ . Perpendicular: $\dfrac23 \cdot \dfrac{\beta - 17}{8} = -1 \Rightarrow \beta = 5$ .
Inclination is measured anticlockwise from the positive $x$ -axis in $[0, \pi)$ : slope $-1$ means $\theta = \dfrac{3\pi}{4}$ , not $-\dfrac{\pi}{4}$ . The line through $(-3, 6)$ and the midpoint $(1, 2)$ of $(4, -5)$ , $(-2, 9)$ has slope $-1$ , inclination $\dfrac{3\pi}{4}$ .
A point $(h, k)$ on $l_1$ through $(1, 2)$ , $(-3, 4)$ (slope $-\tfrac12$ , $x + 2y = 5$ ) such that the line to $(4, 3)$ is perpendicular (slope $2$ , $2x - y = 5$ ): solve, $(3, 1)$ , $\dfrac{k}{h} = \dfrac13$ .
A vertical line has no slope; a horizontal one has slope $0$ . Treat them separately in every formula below.

Slope

m = \tan\theta = \frac{y_2 - y_1}{x_2 - x_1} = -\frac{a}{b} \qquad \text{parallel: } m_1 = m_2 \qquad \text{perpendicular: } m_1 m_2 = -1

Worked example

The line

3x + 4y = 12

is perpendicular to the line through

(1, 2)

and

(4, k)

. Find

k

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q146Easy

Example 1 · Straight Line · Slope, Angle Between Lines and Rotation

The straight line

2x-3y+17=0

is perpendicular to the line passing through the points

(7,17)

and

(15,\beta)

, then

\beta

equals

Reading a negative slope as a negative angle

Inclination lives in

[0, \pi)

. Slope

-1

\dfrac{3\pi}{4}

; option

\dfrac{\pi}{4}

is the sign-blind answer.

Drill 2 more on slope and inclination: parallel means m₁ = m₂, perpendicular means m₁m₂ = −1

Concept 2 of 5: Angle Between Two Lines: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|

The angle between two lines is the difference of their inclinations, and

\tan(\theta_1 - \theta_2)

expands to the formula. The modulus picks the acute angle; if one line is vertical, the angle with a line of slope

m

has

\tan\theta = \left|\dfrac{1}{m}\right|

Definition

$4x - 2y + 13 = 0$ (slope $2$ ) and the equal-intercept line $x + y = a$ (slope $-1$ ): $\tan\theta = \left|\dfrac{-1 - 2}{1 - 2}\right| = 3$ , $\theta = \tan^{-1}3$ .
Normal-form lines $x\cos\alpha + y\sin\alpha = p$ have slope $-\cot\alpha$ , i.e. inclination $\alpha + 90^\circ$ ; the angle between $\alpha = 30^\circ$ and $\alpha = 60^\circ$ is simply $30^\circ$ .
Diagonals of the parallelogram $A(2,-1)$ , $B(0,2)$ , $C(2,3)$ , $D(4,0)$ : $AC$ is vertical, $BD$ has slope $-\tfrac12$ ; $\tan\theta = 2$ , $\theta = \tan^{-1}2$ .
Parallel lines give $\tan\theta = 0$ ; perpendicular ones make the denominator $0$ ( $\theta = 90^\circ$ ).

Angle between lines

\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \qquad \text{one line vertical: } \tan\theta = \left|\frac{1}{m}\right|

Worked example

Find the acute angle between

y = 3x + 1

and

y = \dfrac{x}{2} - 4

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q104Easy

Example 2 · Straight Line · Slope, Angle Between Lines and Rotation

The acute angle between the line

4x- 2y+ 13 = 0

and the line which makes equal intercepts with the co-ordinate axes is

Dropping the modulus and reporting the obtuse angle

\dfrac{m_1 - m_2}{1 + m_1 m_2}

can come out negative; the acute angle uses its absolute value.

\tan^{-1}(-3)

is not an option, but

\tan^{-1}\frac13

— the reciprocal slip — is.

Drill 2 more on angle between two lines: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|

Concept 3 of 5: Lines Through a Point at a Given Angle: Solve the Angle Formula for m

Fix

m_1

and

\tan\theta

, and the angle formula becomes a quadratic in the unknown slope

m

with two roots — one on each side of the given line. Each root with the point gives a line; a root may be

0

(horizontal) or the quadratic may drop to linear, meaning the second line is vertical.

Definition

Through $(3, 2)$ at $45^\circ$ to $x - 2y - 3 = 0$ ( $m_1 = \tfrac12$ ): $\left|\dfrac{m - \frac12}{1 + \frac{m}{2}}\right| = 1 \Rightarrow m = 3$ or $-\tfrac13$ . Lines $3x - y - 7 = 0$ and $x + 3y - 9 = 0$ .
Through $(3, -2)$ at $60^\circ$ to $\sqrt3 x + y = 1$ ( $m_1 = -\sqrt3$ , inclination $120^\circ$ ): inclinations $60^\circ$ and $180^\circ$ , slopes $\sqrt3$ and $0$ . The horizontal one $y = -2$ never meets the $x$ -axis, so the required line is $y - \sqrt3 x + 2 + 3\sqrt3 = 0$ .
Thinking in inclinations is faster than the quadratic: the two lines have inclinations $\theta_1 \pm \alpha$ .
Check any extra condition in the stem (meets the $x$ -axis; passes through a quadrant) — it is there to kill one of the two roots.

Two slopes

\left|\frac{m - m_1}{1 + m m_1}\right| = \tan\alpha \ \Rightarrow\ m = \frac{m_1 \pm \tan\alpha}{1 \mp m_1\tan\alpha} \quad(\text{inclinations } \theta_1 \pm \alpha)

Worked example

Find the lines through

(1, 1)

making

45^\circ

with

y = 2x

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q142Moderate

Example 3 · Straight Line · Slope, Angle Between Lines and Rotation

The equations of the lines passing through the point

(3, 2)

and making an acute angle of 45° with the line

x - 2y - 3 = 0

are

Keeping only one root

The angle condition always yields TWO lines. Option lists pair the right line with a wrong partner (a sign flipped in the second equation); check both.

Drill 1 more on lines through a point at a given angle: solve the angle formula for m

Concept 4 of 5: Rotating a Line About a Point: Add or Subtract the Angle From the Inclination

Rotation about a point on the line keeps that point and changes the inclination by the rotation angle — anticlockwise adds, clockwise subtracts. The new slope is

\tan

of the new inclination, and the line is written through the fixed point.

Definition

$AB$ through $A(2, 0)$ , $B(3, 1)$ has inclination $45^\circ$ ; rotated anticlockwise by $15^\circ$ it is at $60^\circ$ : $y = \sqrt3(x - 2)$ .
A line through $A(2, 0)$ at $30^\circ$ rotated CLOCKWISE by $15^\circ$ is at $15^\circ$ : $y = (2 - \sqrt3)(x - 2)$ , i.e. $(2 - \sqrt3)x - y - 4 + 2\sqrt3 = 0$ .
$x - y - 2 = 0$ meets the $x$ -axis at $M(2, 0)$ with inclination $45^\circ$ ; rotated anticlockwise by $45^\circ$ it is vertical: $x = 2$ .
$\tan 15^\circ = 2 - \sqrt3$ , $\tan 75^\circ = 2 + \sqrt3$ ; $90^\circ$ means a vertical line, not an infinite slope to substitute.

Rotation

\theta_{\text{new}} = \theta \pm \alpha \ (+\text{ anticlockwise}),\qquad y - y_0 = \tan\theta_{\text{new}}\,(x - x_0)

Worked example

The line through

P(1, 2)

with inclination

60^\circ

is rotated clockwise about

P

through

30^\circ

. Find the new line.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q135Moderate

Example 4 · Straight Line · Slope, Angle Between Lines and Rotation

If the line joining two points

A(2,0)

and

B(3,1)

is rotated about

A

in anticlockwise direction through an angle of

15^\circ

, then the equation of the line in the new position is

Rotating the wrong way

Clockwise subtracts.

30^\circ - 15^\circ = 15^\circ

gives slope

2 - \sqrt3

; adding gives

45^\circ

and a line that is on the list as a distractor.

Drill 2 more on rotating a line about a point: add or subtract the angle from the inclination

Concept 5 of 5: Bisector of an Angle at a Vertex, and the Reflection of a Line

At a vertex

Q

, the bisector of

\angle PQR

points along the average of the directions

QP

and

QR

. Reflecting a line in another keeps the angle between them, so the reflected slope is the second root of the angle formula — the same trick as the equal-distance stem.

Definition

$P(-5, 0)$ , $Q(0, 0)$ , $R(2, 2\sqrt3)$ : $QP$ points at $180^\circ$ , $QR$ at $60^\circ$ ; the bisector is at $120^\circ$ , slope $-\sqrt3$ : $\sqrt3 x + y = 0$ . The 2024 twin with $R(3, 3\sqrt3)$ is the same picture.
$A(2, -7)$ ; $AB$ is $4x + y = 1$ (slope $-4$ ), $BC$ is $3x - 4y + 1 = 0$ (slope $\tfrac34$ ); $AB = AC$ means $AC$ is the reflection of $AB$ in the perpendicular from $A$ to $BC$ , equivalently the other line through $A$ making the same angle with $BC$ : $\dfrac{m - \frac34}{1 + \frac{3m}{4}} = -\dfrac{19}{8} \Rightarrow m = -\dfrac{52}{89}$ ; $52x + 89y + 519 = 0$ .
Distances from $(2, 5)$ to $3x + y + 4 = 0$ measured along two lines are equal iff the two lines make equal angles with it: slope $\tfrac34$ makes $\tan\theta = 3$ ; the other slope with $\tan\theta = 3$ is $m = 0$ .
Bisector direction as an average works only with directions measured at the vertex; compute both direction angles from $Q$ , not the lines' inclinations.

Bisector direction

\theta_{\text{bisector}} = \frac{\theta_{QP} + \theta_{QR}}{2} \qquad \text{reflected slope: the other root of } \left|\frac{m - m_L}{1 + m m_L}\right| = \tan\theta

Worked example

Find the bisector of

\angle AOB

where

O

is the origin,

A(4, 0)

and

B(0, 4)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q146Hard

Example 5 · Straight Line · Slope, Angle Between Lines and Rotation

Let

P \equiv (-5,0)

Q \equiv (0,0)

and

R \equiv (2,2\sqrt{3})

be three points. Then the equation of the bisector of the angle PQR is

Bisecting the inclinations instead of the directions

QP

has inclination

0^\circ

as a line but direction

180^\circ

from

Q

. Averaging

0^\circ

and

60^\circ

gives

30^\circ

— the bisector of the OTHER angle at

Q

, and its equation is offered.

Drill 3 more on bisector of an angle at a vertex, and the reflection of a line

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Slope and Inclination: Parallel Means m₁ = m₂, Perpendicular Means m₁m₂ = −1
Slope
$m = \tan\theta = \frac{y_2 - y_1}{x_2 - x_1} = -\frac{a}{b} \qquad \text{parallel: } m_1 = m_2 \qquad \text{perpendicular: } m_1 m_2 = -1$
Angle Between Two Lines: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|
Angle between lines
$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \qquad \text{one line vertical: } \tan\theta = \left|\frac{1}{m}\right|$
Lines Through a Point at a Given Angle: Solve the Angle Formula for m
Two slopes
$\left|\frac{m - m_1}{1 + m m_1}\right| = \tan\alpha \ \Rightarrow\ m = \frac{m_1 \pm \tan\alpha}{1 \mp m_1\tan\alpha} \quad(\text{inclinations } \theta_1 \pm \alpha)$
Rotating a Line About a Point: Add or Subtract the Angle From the Inclination
Rotation
$\theta_{\text{new}} = \theta \pm \alpha \ (+\text{ anticlockwise}),\qquad y - y_0 = \tan\theta_{\text{new}}\,(x - x_0)$
Bisector of an Angle at a Vertex, and the Reflection of a Line
Bisector direction
$\theta_{\text{bisector}} = \frac{\theta_{QP} + \theta_{QR}}{2} \qquad \text{reflected slope: the other root of } \left|\frac{m - m_L}{1 + m m_L}\right| = \tan\theta$

Watch out for (5)

Reading a negative slope as a negative angle
→ Slope and Inclination: Parallel Means m₁ = m₂, Perpendicular Means m₁m₂ = −1
Dropping the modulus and reporting the obtuse angle
→ Angle Between Two Lines: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|
Keeping only one root
→ Lines Through a Point at a Given Angle: Solve the Angle Formula for m
Rotating the wrong way
→ Rotating a Line About a Point: Add or Subtract the Angle From the Inclination
Bisecting the inclinations instead of the directions
→ Bisector of an Angle at a Vertex, and the Reflection of a Line

Test yourself on Straight Line

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Related notes

Pair of Straight Lines — the same angle formula for a joint equation →