Class 11 Applied Mathematics

Straight
Lines

The intersection of algebra and geometry. From fundamental distances to complex normal forms and coordinate transformations.

Foundational Geometry

Coordinates & Ratios

Distance Formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Derived from the Pythagorean theorem: $a^2 + b^2 = c^2$.

Section & Centroid

Centroid ($G$)

G(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)

Internal Division

P(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)

Coordinate Mapping

Visualize distance and section formulas on a cartesian plane.

A (x1, y1)

B (x2, y2)

d \approx \sqrt{(\Delta x)^2 + (\Delta y)^2}

Logic

Points are processed as $(x, y)$ coordinates for all calculations.

Section Formula

Ratios $(m:n)$ determine position between endpoints.

The General Key: $Ax + By + C = 0$

Slope (m)

$-A/B$

Vertical Ratio

Y-Intercept

$-C/B$

Intersection @ Y

X-Intercept

$-C/A$

Intersection @ X

Tip: Always convert non-standard equations to General Form before extraction.

Slope & Trigonometric Relation

Slope & Inclination

Observe how the angle of inclination $\theta$ affects the slope $m = \tan \theta$.

Q2 ($90-180^\circ$)Q1 ($0-90^\circ$)

Angle of Inclination: 45°

Calculation

\tan(45^\circ)

Result (m)

1.000

Q2 Reduction

\tan(180^\circ-135^\circ) = -\tan(135^\circ)

Cot Relation

\tan(90^\circ+0^\circ) = -\cot(0^\circ)

For

\theta

between

90^\circ

and

180^\circ

, the slope is **negative** because

\tan \theta < 0

in the second quadrant.

Wave Functions

Motion of Sin and Cos from $0^\circ \to 180^\circ$

0°

Sin

0

(0.00)

Cos

1

(1.00)

30°

Sin

1/2

(0.50)

Cos

\sqrt{3}/2

(0.87)

45°

Sin

1/\sqrt{2}

(0.71)

Cos

1/\sqrt{2}

(0.71)

60°

Sin

\sqrt{3}/2

(0.87)

Cos

1/2

(0.50)

90°

Sin

1

(1.00)

Cos

0

(0.00)

120°

Sin

\sqrt{3}/2

(0.87)

Cos

-1/2

(-0.50)

135°

Sin

1/\sqrt{2}

(0.71)

Cos

-1/\sqrt{2}

(-0.71)

150°

Sin

1/2

(0.50)

Cos

-\sqrt{3}/2

(-0.87)

180°

Sin

0

(0.00)

Cos

-1

(-1.00)

The Trig Atlas

Essential values from 0° to 180° for slope calculations.

ASTC RuleOnly Sin is +ve in Q2 (90-180)

Angle ($\theta$)	Sin $\theta$	Cos $\theta$	Tan $\theta$ (Slope)
0°	0	1	$0$
30°	1/2	√3/2	$1/\sqrt3$
45°	√2/2	√2/2	$1$
60°	√3/2	1/2	$\sqrt3$
90°	1	0	$\infty$
120°	√3/2	-1/2	$-\sqrt3$
135°	√2/2	-√2/2	$-1$
150°	1/2	-√3/2	$-1/\sqrt3$
180°	0	-1	$0$

Forms of Equation

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 = mx + c

intercept Form

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

normal Form

x \cos \alpha + y \sin \alpha = p

general Form

Ax + By + C = 0

Angle Between Lines

The acute angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is determined by the tangent formula.

\tan \theta = \left \mid \frac{m_2 - m_1}{1 + m_1 m_2} \right|

Parallel: $m_1 = m_2$

Perpendicular: $m_1 m_2 = -1$

Compound Trig

\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

Distances & Proximity

Point to Line

The perpendicular distance $d$ from a point $P(x_1, y_1)$ to the line $Ax + By + C = 0$.

d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Parallel Lines

The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$.

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

Practice Laboratory

Geometric Workouts

8 Challenge Problems

CBSE Ready

Analysis Zone

Find the coordinates of the centroid of a triangle whose vertices are (1, 4), (2, 7) and (-3, -2).

✓ Show Solution

1. Vertices:

(1, 4), (2, 7), (-3, -2)

2. Centroid

G = (\frac{1+2-3}{3}, \frac{4+7-2}{3}) = (\frac{0}{3}, \frac{9}{3})

G = (0, 3)

EASY

Find the slope of a line which makes an angle of 150° with the positive direction of x-axis.

✓ Show Solution

\theta = 150^\circ

m = \tan(150^\circ) = \tan(180^\circ - 30^\circ) = -\tan(30^\circ)

m = -1/\sqrt{3}

EASY

Find the equation of a line passing through (2, 3) and perpendicular to the line 4x + 3y + 10 = 0.

✓ Show Solution

1. Given line:

4x + 3y + 10 = 0 \implies m_1 = -4/3

2. Perpendicular slope

m_2 = -1/m_1 = 3/4

3. Equation:

y - 3 = \frac{3}{4}(x - 2) \implies 4y - 12 = 3x - 6

4. Final Form:

3x - 4y + 6 = 0

MEDIUM

Calculate the distance between the parallel lines 3x - 4y + 7 = 0 and 3x - 4y + 5 = 0.

✓ Show Solution

A=3, B=-4, C_1=7, C_2=5

d = \frac{|7 - 5|}{\sqrt{3^2 + (-4)^2}} = \frac{2}{\sqrt{9 + 16}} = \frac{2}{5} = 0.4

units

MEDIUM

Find the value of k if the line (k-3)x - (4-k²)y + (k²-7k+6) = 0 is parallel to x-axis.

✓ Show Solution

1. Line parallel to x-axis means slope

m = 0

m = -\frac{k-3}{-(4-k^2)} = 0 \implies k-3 = 0

3. Result:

k = 3

HARD

Reduction to Normal Form: Reduce x - √3y + 8 = 0 into normal form.

✓ Show Solution

x - \sqrt{3}y = -8 \implies -x + \sqrt{3}y = 8

2. Divide by

\sqrt{(-1)^2 + (\sqrt{3})^2} = 2

-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 4

\cos(120^\circ)x + \sin(120^\circ)y = 4

HARD

Find the distance of the point (3, -5) from the line 3x - 4y - 26 = 0.

✓ Show Solution

1. Point

(x_1, y_1) = (3, -5)

, Line

3x - 4y - 26 = 0

d = \frac{|3(3) - 4(-5) - 26|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}}

d = \frac{|3|}{5} = 0.6

units

MEDIUM

Find the distance between the parallel lines

15x + 8y - 34 = 0

and

15x + 8y + 31 = 0

✓ Show Solution

A = 15, B = 8, C_1 = -34, C_2 = 31

d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}} = \frac{|-65|}{\sqrt{225 + 64}}

d = \frac{65}{\sqrt{289}} = \frac{65}{17}

units

MEDIUM

Competitive Corner

Board & Previous Year Questions

CBSE 2023

Find the equation of the line passing through the intersection of 2x + 3y = 4 and 3x - y = 5 and is perpendicular to x+3y=7.

Official Solution1. Solve intersection:

x=19/11, y=2/11

(approx) 2. Perp slope to

x+3y=7

m=3

3. Use point-slope form:

y - 2/11 = 3(x - 19/11)

11y - 2 = 33x - 57 \implies 33x - 11y - 55 = 0

CBSE 2021

Find the ratio in which the line 3x + 4y = 7 divides the segment joining (1, 2) and (-2, 1).

Official Solution1. Let ratio be

k:1

. Point

P = (\frac{-2k+1}{k+1}, \frac{k+2}{k+1})

P

lies on

3x+4y=7 \implies 3(\frac{-2k+1}{k+1}) + 4(\frac{k+2}{k+1}) = 7

-6k + 3 + 4k + 8 = 7k + 7 \implies -2k + 11 = 7k + 7

9k = 4 \implies k = 4/9

. Ratio is

4:9

Board Favorite

Find the equation of a line which passes through (2, 2) and whose sum of intercepts on the axes is 9.

Official Solution1.

a + b = 9 \implies b = 9 - a

\frac{x}{a} + \frac{y}{9-a} = 1

. Passes through (2, 2): 3.

\frac{2}{a} + \frac{2}{9-a} = 1 \implies 2(9-a) + 2a = a(9-a) \implies 18 = 9a - a^2

a^2 - 9a + 18 = 0 \implies (a-3)(a-6) = 0

. Lines:

x/3 + y/6 = 1

and

x/6 + y/3 = 1

Proof Based

Distance from Origin: A line is such that its segment between the axes is bisected at the point (h, k). Show that its equation is x/h + y/k = 2.

Official Solution1. Let intercepts be

(a, 0)

and

(0, b)

. 2. Midpoint:

(a/2, b/2) = (h, k) \implies a = 2h, b = 2k

3. Intercept form:

\frac{x}{2h} + \frac{y}{2k} = 1

4. Multiply by 2:

\frac{x}{h} + \frac{y}{k} = 2

Advanced CBSE

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Official Solution1. Line:

x + 3y - 7 = 0

. Point:

(3, 8)

. 2. Mirror formula:

\frac{x-3}{1} = \frac{y-8}{3} = -2 \frac{(3 + 3(8) - 7)}{1^2 + 3^2}

\frac{x-3}{1} = \frac{y-8}{3} = -2 \frac{20}{10} = -4

x = 3-4 = -1, y = 8-12 = -4

. Image is

(-1, -4)

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Class 11 Applied Mathematics

Straight
Lines

The intersection of algebra and geometry. From fundamental distances to complex normal forms and coordinate transformations.

Foundational Geometry

Coordinates & Ratios

Distance Formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Derived from the Pythagorean theorem: $a^2 + b^2 = c^2$.

Section & Centroid

Centroid ($G$)

G(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)

Internal Division

P(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)

Coordinate Mapping

Visualize distance and section formulas on a cartesian plane.

A (x1, y1)

B (x2, y2)

d \approx \sqrt{(\Delta x)^2 + (\Delta y)^2}

Logic

Points are processed as $(x, y)$ coordinates for all calculations.

Section Formula

Ratios $(m:n)$ determine position between endpoints.

The General Key: $Ax + By + C = 0$

Slope (m)

$-A/B$

Vertical Ratio

Y-Intercept

$-C/B$

Intersection @ Y

X-Intercept

$-C/A$

Intersection @ X

Tip: Always convert non-standard equations to General Form before extraction.

Slope & Trigonometric Relation

Slope & Inclination

Observe how the angle of inclination $\theta$ affects the slope $m = \tan \theta$.

Q2 ($90-180^\circ$)Q1 ($0-90^\circ$)

Angle of Inclination: 45°

Calculation

\tan(45^\circ)

Result (m)

1.000

Q2 Reduction

\tan(180^\circ-135^\circ) = -\tan(135^\circ)

Cot Relation

\tan(90^\circ+0^\circ) = -\cot(0^\circ)

For

\theta

between

90^\circ

and

180^\circ

, the slope is **negative** because

\tan \theta < 0

in the second quadrant.

Wave Functions

Motion of Sin and Cos from $0^\circ \to 180^\circ$

0°

Sin

0

(0.00)

Cos

1

(1.00)

30°

Sin

1/2

(0.50)

Cos

\sqrt{3}/2

(0.87)

45°

Sin

1/\sqrt{2}

(0.71)

Cos

1/\sqrt{2}

(0.71)

60°

Sin

\sqrt{3}/2

(0.87)

Cos

1/2

(0.50)

90°

Sin

1

(1.00)

Cos

0

(0.00)

120°

Sin

\sqrt{3}/2

(0.87)

Cos

-1/2

(-0.50)

135°

Sin

1/\sqrt{2}

(0.71)

Cos

-1/\sqrt{2}

(-0.71)

150°

Sin

1/2

(0.50)

Cos

-\sqrt{3}/2

(-0.87)

180°

Sin

0

(0.00)

Cos

-1

(-1.00)

The Trig Atlas

Essential values from 0° to 180° for slope calculations.

ASTC RuleOnly Sin is +ve in Q2 (90-180)

Angle ($\theta$)	Sin $\theta$	Cos $\theta$	Tan $\theta$ (Slope)
0°	0	1	$0$
30°	1/2	√3/2	$1/\sqrt3$
45°	√2/2	√2/2	$1$
60°	√3/2	1/2	$\sqrt3$
90°	1	0	$\infty$
120°	√3/2	-1/2	$-\sqrt3$
135°	√2/2	-√2/2	$-1$
150°	1/2	-√3/2	$-1/\sqrt3$
180°	0	-1	$0$

Forms of Equation

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 = mx + c

intercept Form

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

normal Form

x \cos \alpha + y \sin \alpha = p

general Form

Ax + By + C = 0

Angle Between Lines

The acute angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is determined by the tangent formula.

\tan \theta = \left \mid \frac{m_2 - m_1}{1 + m_1 m_2} \right|

Parallel: $m_1 = m_2$

Perpendicular: $m_1 m_2 = -1$

Compound Trig

\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

Distances & Proximity

Point to Line

The perpendicular distance $d$ from a point $P(x_1, y_1)$ to the line $Ax + By + C = 0$.

d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Parallel Lines

The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$.

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

Practice Laboratory

Geometric Workouts

8 Challenge Problems

CBSE Ready

Analysis Zone

Find the coordinates of the centroid of a triangle whose vertices are (1, 4), (2, 7) and (-3, -2).

✓ Show Solution

1. Vertices:

(1, 4), (2, 7), (-3, -2)

2. Centroid

G = (\frac{1+2-3}{3}, \frac{4+7-2}{3}) = (\frac{0}{3}, \frac{9}{3})

G = (0, 3)

EASY

Find the slope of a line which makes an angle of 150° with the positive direction of x-axis.

✓ Show Solution

\theta = 150^\circ

m = \tan(150^\circ) = \tan(180^\circ - 30^\circ) = -\tan(30^\circ)

m = -1/\sqrt{3}

EASY

Find the equation of a line passing through (2, 3) and perpendicular to the line 4x + 3y + 10 = 0.

✓ Show Solution

1. Given line:

4x + 3y + 10 = 0 \implies m_1 = -4/3

2. Perpendicular slope

m_2 = -1/m_1 = 3/4

3. Equation:

y - 3 = \frac{3}{4}(x - 2) \implies 4y - 12 = 3x - 6

4. Final Form:

3x - 4y + 6 = 0

MEDIUM

Calculate the distance between the parallel lines 3x - 4y + 7 = 0 and 3x - 4y + 5 = 0.

✓ Show Solution

A=3, B=-4, C_1=7, C_2=5

d = \frac{|7 - 5|}{\sqrt{3^2 + (-4)^2}} = \frac{2}{\sqrt{9 + 16}} = \frac{2}{5} = 0.4

units

MEDIUM

Find the value of k if the line (k-3)x - (4-k²)y + (k²-7k+6) = 0 is parallel to x-axis.

✓ Show Solution

1. Line parallel to x-axis means slope

m = 0

m = -\frac{k-3}{-(4-k^2)} = 0 \implies k-3 = 0

3. Result:

k = 3

HARD

Reduction to Normal Form: Reduce x - √3y + 8 = 0 into normal form.

✓ Show Solution

x - \sqrt{3}y = -8 \implies -x + \sqrt{3}y = 8

2. Divide by

\sqrt{(-1)^2 + (\sqrt{3})^2} = 2

-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 4

\cos(120^\circ)x + \sin(120^\circ)y = 4

HARD

Find the distance of the point (3, -5) from the line 3x - 4y - 26 = 0.

✓ Show Solution

1. Point

(x_1, y_1) = (3, -5)

, Line

3x - 4y - 26 = 0

d = \frac{|3(3) - 4(-5) - 26|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}}

d = \frac{|3|}{5} = 0.6

units

MEDIUM

Find the distance between the parallel lines

15x + 8y - 34 = 0

and

15x + 8y + 31 = 0

✓ Show Solution

A = 15, B = 8, C_1 = -34, C_2 = 31

d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}} = \frac{|-65|}{\sqrt{225 + 64}}

d = \frac{65}{\sqrt{289}} = \frac{65}{17}

units

MEDIUM

Competitive Corner

Board & Previous Year Questions

CBSE 2023

Find the equation of the line passing through the intersection of 2x + 3y = 4 and 3x - y = 5 and is perpendicular to x+3y=7.

Official Solution1. Solve intersection:

x=19/11, y=2/11

(approx) 2. Perp slope to

x+3y=7

m=3

3. Use point-slope form:

y - 2/11 = 3(x - 19/11)

11y - 2 = 33x - 57 \implies 33x - 11y - 55 = 0

CBSE 2021

Find the ratio in which the line 3x + 4y = 7 divides the segment joining (1, 2) and (-2, 1).

Official Solution1. Let ratio be

k:1

. Point

P = (\frac{-2k+1}{k+1}, \frac{k+2}{k+1})

P

lies on

3x+4y=7 \implies 3(\frac{-2k+1}{k+1}) + 4(\frac{k+2}{k+1}) = 7

-6k + 3 + 4k + 8 = 7k + 7 \implies -2k + 11 = 7k + 7

9k = 4 \implies k = 4/9

. Ratio is

4:9

Board Favorite

Find the equation of a line which passes through (2, 2) and whose sum of intercepts on the axes is 9.

Official Solution1.

a + b = 9 \implies b = 9 - a

\frac{x}{a} + \frac{y}{9-a} = 1

. Passes through (2, 2): 3.

\frac{2}{a} + \frac{2}{9-a} = 1 \implies 2(9-a) + 2a = a(9-a) \implies 18 = 9a - a^2

a^2 - 9a + 18 = 0 \implies (a-3)(a-6) = 0

. Lines:

x/3 + y/6 = 1

and

x/6 + y/3 = 1

Proof Based

Distance from Origin: A line is such that its segment between the axes is bisected at the point (h, k). Show that its equation is x/h + y/k = 2.

Official Solution1. Let intercepts be

(a, 0)

and

(0, b)

. 2. Midpoint:

(a/2, b/2) = (h, k) \implies a = 2h, b = 2k

3. Intercept form:

\frac{x}{2h} + \frac{y}{2k} = 1

4. Multiply by 2:

\frac{x}{h} + \frac{y}{k} = 2

Advanced CBSE

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Official Solution1. Line:

x + 3y - 7 = 0

. Point:

(3, 8)

. 2. Mirror formula:

\frac{x-3}{1} = \frac{y-8}{3} = -2 \frac{(3 + 3(8) - 7)}{1^2 + 3^2}

\frac{x-3}{1} = \frac{y-8}{3} = -2 \frac{20}{10} = -4

x = 3-4 = -1, y = 8-12 = -4

. Image is

(-1, -4)