## Angle between two lines

Subject: Optional Mathematics

#### Overview

The angle between two lines

Angle between the lines y = m1 + c1 and y = m2 + c2 and y = m2x + c2

Condition of perpendicularity

m1m2 = -1

Condition of Parallelism

m1 = m2

Angle between the lines A1x + B1y + C1 = 0 and A2x + B2y +C2 = 0

Condition of perpendicularity

A1A2 + B1B2 = 0

Condition of Parallelism

A1B2 = A2B1

Equation of any line parallel to ax + by + c = 0

k = -bc

Equation of any line perpendicular to ax +by +c = 0

k = ac

#### Angle between the lines y = m1 + c1 and y = m2 + c2 and y = m2x + c2

Let the equation of two lines AB and CD be y = m1x + c1 and y = m2x + c2 respectively.

let the lines AB and CD make angles θ1 and θ2 respectively with the positive direction of X-axis.

Then, tanθ1 = m1 and tanθ2 = m2.

Let the lines AB and CD intersect each other at the point E.

Let the angles between the lines AB and CD

∠CEA = Φ

Then by plane geometry, θ1 = Φ + θ2

or,Φ =θ12

∴ tanΦ = tan(θ1 2) = $\frac {tan\theta_1 - tan\theta_2}{1 + tan\theta_1 tan\theta_2}$ = $\frac{m_1 - m_2}{1+tan\theta_1 tan\theta_2}$ .........(i)

Again, let ∠BAC = Ψ

Then by place geometry ,Φ + Ψ = 1800

or, Ψ = 1800 - Φ

or, tan Ψ = tan (180 - Φ) = -tan Φ = - $\frac{m_1 - m_2}{1+m_1m_2}$ ............(ii)

Hence if angles between the lines y = m1x + c1 and y = m2x + c2 be the θ then,

tanθ =± $\frac{m_1 - m_2}{1+m_1m_2}$

θ = tan-1(± $\frac{m_1 - m_2}{1+m_1m_2}$)

Condition of Perpendicularity

Two lines AB and CD will be perpendicular to each other if the angle between them θ = 90o.

We have tanθ = ± $\frac{m_1 - m_2}{1+m_1m_2}$

or, tan900 = ± $\frac{m_1 - m_2}{1+m_1m_2}$

or, cot900 = ± $\frac{1+m_1m_2}{m_1 - m_2}$

or, 0 = $\frac{1+m_1m_2}{m_1 - m_2}$

or, 1 +m1m2 =0

or, m1m2 = -1

Two lines will be perpendicular to each other if m1m2 = -1

i.e. if product of the slopes = -1 .

Condition of Parallelism

Two lines AB and CD will be parallel to each other if the angle between them θ = 00.

we have, tanθ = ± $\frac{m_1 - m_2}{1+m_1m_2}$

or, tan00 = ± $\frac{m_1 - m_2}{1+m_1m_2}$

or, 0 = $\frac{m_1 - m_2}{1+m_1m_2}$

or, m1 - m2 = 0

or, m1 = m2

∴ Two lines will be parallel to each other if m1 = m2 i.e. if slopes are equal.

#### Angle between the lines A1x + B1y + C1 = 0 and A2x + B2y +C2 = 0

Let equations of two straight lines AB and CD be A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0 respectively.

Then slope of AB = -$\frac{A_1}{B_1}$

Slope of CD = -$\frac{A_2}{B_2}$

Let the lines AB and CD make angles θ1 and θ2 with the positive direction of X-axis.

Then, tanθ1 = -$\frac{A_1}{B_1}$ and tanθ2 = -$\frac{A_2}{B_2}$

Let ∠CEA = Φ.

Then, θ1 = θ1 - θ2

or, Φ = θ1 - θ2

∴ tan Φ = tan( θ1 - θ2) = $\frac{tan\theta_1 - tan\theta_2}{1 + tan\theta_1 tan\theta_2}$ = $\frac{A_2 B_1 - A_1 B_2}{A_1 A_2 + B_1 B_2}$ = -$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$ ......(i)

Let ∠BEC = Ψ

Then Ψ + Φ = 1800

or, Ψ = 1800 - Φ

∴ tan Ψ = tan(1800 - Φ) = -tanΦ = $\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$ .........(ii)

Hence if angles between the lines A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0 is θ, then

tanθ = ± $\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$

or, θ = tan-1 (±$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$ )

Condition of Perpendicularity

Two lines AB and CD will be perpendicular to each other if θ = 900

Then tan900 = ± $\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$

or, ∞ = $\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$

∴ A1A2 + B1B2 = 0

Condition of Parallelism

Two lines AB and CD will be parrallel to each other if θ = 00

Then, tan00 = ± $\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$

or, 0 = $\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$

or, A1B2 - A2B1 = 0

or, A1B2 = A2B1

∴ $\frac{A_1}{A_2}$ = $\frac{B_1}{B_2}$

#### Equation of any line parallel to ax + by + c = 0

Equation of the given line is ax + by + c = 0

Slope of this line = -$\frac{coefficient \;of \;x}{coefficient \;of \;y}$ = -$\frac{a}{b}$

Slope of the line parallel to this line = -$\frac{a}{b}$

Now, equation of a line having slope -$\frac{a}{b}$ is given by

y = mx + c

or, y = -$\frac{a}{b}$x + c

or, by = -ax + bc

or, ax + by - bc = 0

or, ax + by + k = 0 where, k = -bc.

Hence equation of any line parallel to ax + by + c = 0 is given by ax + by + k = 0 where k is an arbitrary constant.

#### Equation of any line perpendicular to ax +by +c = 0

Equation of the given line is ax + by + c = 0

Slope of this line = -$\frac{coefficient\;of\;x}{coefficient\;of\;y}$ = -$\frac{a}{b}$

Slope of the line perpendicular to given line = $\frac{b}{a}$

Now equation of a line having slope $\frac{b}{a}$ is given by

y = mx + c

or, y = $\frac{b}{a}$x + c

or, ay = bx + ac

or, bx - ay + ac = 0

or, bx - ay +k = 0 where k = ac.

Hence, equation of any line perpendicular to ax + by + c = 0 is given by bx - ay + k = 0 where k is an arbitrary constant.

##### Things to remember

Angle between the lines y = m1 + c1 and y = m2 + c2 and y = m2x + c2

Condition of perpendicularity

m1m2 = -1

Condition of Parallelism

m1 = m2

Angle between the lines A1x + B1y + C1 = 0 and A2x + B2y +C2 = 0

Condition of perpendicularity

A1A2 + B1B2 = 0

Condition of Parallelism

A1B2 = A2B1

Equation of any line parallel to ax + by + c = 0

k = -bc

Equation of any line perpendicular to ax +by +c = 0

k = ac

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##### Vectors : Angle between two lines given their equations

Given:

a1x + b1y + c1 = 0..............................(1)

a2x + b2y + c2 = 0..............................(2)

Slope of eqn (1), m1 = - $\frac {x-coefficient}{y-coefficient}$ = -$\frac {a_1}{b_1}$

Slope of eqn (2), m2 = - $\frac {x-coefficient}{y-coefficient}$ = -$\frac {a_2}{b_2}$

when the lines are parallel, then:

m1 = m2

or,-$\frac {a_1}{b_1}$ =-$\frac {a_2}{b_2}$

∴ a1b2 = a2b1 Ans

when the lines are perpendicular, then:

m1× m2= - 1

or,-$\frac {a_1}{b_1}$×-$\frac {a_2}{b_2}$ = - 1

∴ a1a2 = - b1b2Ans

Here,

5x + 4y - 10 = 0............................(1)

15x + 12y - 7 = 0.........................(2)

Slope of eqn (1), m1 = - $\frac {x-coefficient}{y-coefficient}$ = -$\frac 54$

Slope of eqn (2), m2 = - $\frac {x-coefficient}{y-coefficient}$ = -$\frac {15}{12}$ = -$\frac 54$

m1 = m2 = - $\frac 54$

∴ The given two lines are parallel to each other. Proved

Here,

x + 3y = 2..................................(1)

6x - 2y = 9................................(2)

Slope of eqn (1), m1 = - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 13$

Slope of eqn (2), m2 = - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 6{-2}$ =3

We have,

m1× m2 =- $\frac 13$× 3 = - 1

∴m1× m2 = - 1

Hence, the given two lines are perpendicular each other. Proved

Here,

3x - 2y - 5 = 0..............................(1)

2x + py - 3 = 0.............................(2)

Slope of eqn (1), m1 = - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 3{-2}$ = $\frac 32$

Slope of eqn (2), m2 = - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 2{p}$

When two lines are parallel lines, then:

m1 = m2

or,$\frac 32$ = - $\frac 2{p}$

or, p = $\frac {-2 × 2}3$

∴ p = -$\frac 43$ Ans

Here,

4x + ky - 4 = 0..............................(1)

2x - 6y = 5.............................(2)

Slope of eqn (1), m1 = - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 4k$

Slope of eqn (2), m2 = - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 2{-6}$ = $\frac 13$

when two line are perpendicular to each other,

m1× m2 = - 1

or, - $\frac 4k$ × $\frac 13$ = - 1

or, - $\frac 43$× - 1 = k

∴ k = $\frac 43$ Ans

The formulae of angle between y = m1x + c1 andy = m2x + c2is:

tan$\theta$ = ± $\frac {m_1 - m_2}{1 + m_1m_2}$

when m1× m2 = -1, the two lines are perpendicular to each other.

when m1 = m2, the two lines are parallel to each other.

Here,

Slope of points (3, -4) and (-2, a)

m1 = $\frac {y_2 - y_1}{x_2 - x_1}$ = $\frac {a + 4}{-2 - 3}$ = $\frac {-(a + 4)}5$

Given eqn is y + 2x + 3 = 0

Slope of above eqn (m2) = - $\frac {x-coefficient}{y-coefficient}$ = -$\frac 21$ = - 2

when lines are parallel then,

m1 = m2

or,$\frac {-(a + 4)}5$ = - 2

or, a + 4 = 10

or, a = 10 - 4

∴ a = 6 Ans

Here,

Slope of the points (3, -4) and (-2, 6) is:

m1 = $\frac {y_2 - y_1}{x_2 - x_1}$ = $\frac {6 + 4}{-2 - 3}$ = $\frac {10}{-5}$ = -2

Slope of the eqn y + 2x + 3 = 0 is:

m2 = - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 21$ = -2

From above,

m1= m2 = - 2

Hence, the lines are parallel. Proved

Here,

Given eqn is kx - 3y + 6 = 0

Slope of above eqn is:

m1 = - $\frac {x-coefficient}{y-coefficient}$ = - $\frac k{-3}$ = $\frac k3$

Slope ofthe point (4, 3) and (5, -3) is:

m2 = $\frac {y_2 - y_1}{x_2 - x_1}$ = $\frac {-3 - 3}{5 - 4}$ = - $\frac 61$ = -6

If lines are perpendicular then:

m1× m2 = -1

or, $\frac k3$× -6 = -1

or, k = $\frac {-1}{-2}$

∴ k = $\frac 12$ Ans

Here,

Given equations of the lines are:

y = m1x + c1...............................(1)

y = m2x + c2...............................(2)

If $\theta$ be the angle between two lines (1) and (2);

The formula of angle between the given lines is:

tan$\theta$ =± $\frac {m_1 - m_2}{1 + m_1m_2}$

∴ $\theta$ = tan-1(± $\frac {m_1 - m_2}{1 + m_1m_2}$)

If two lines are perpendicular ($\theta$ = 90°)

tan 90° =± $\frac {m_1 - m_2}{1 + m_1m_2}$

or,∞=± $\frac {m_1 - m_2}{1 + m_1m_2}$

or, $\frac 10$ =± $\frac {m_1 - m_2}{1 + m_1m_2}$

or, 1 + m1m2 = 0

∴ m1m2 = -1 Ans

Here,

The given equations are:

2x + ay + 3 = 0....................(1)

3x - 2y = 5.............................(2)

Slope of equation (1), m1 = -$\frac {x-coefficient}{y-coefficient}$ = - $\frac 2a$

Slope of equation (2), m2 = -$\frac {x-coefficient}{y-coefficient}$ = - $\frac 3{-2}$ = $\frac 32$

If equation (1) and equation (2) are perpendicular to each other:

m1× m2 = -1

or,- $\frac 2a$×$\frac 32$ = -1

or, -6 = - 2a

or, a = $\frac 62$

∴ a = 3 Ans

Here,

Given equations of the lines are:

2x + 4y - 7 = 0...........................(1)

6x + 12y + 4 = 0.......................(2)

Slope of equation (1) is: m1 = - $\frac {x-coefficient}{y-coefficient}$ = -$\frac 24$ = -$\frac 12$

Slope of equation (2) is: m2 = - $\frac {x-coefficient}{y-coefficient}$ = -$\frac 6{12}$ = -$\frac 12$

∴ m1 = m2 = $\frac {-1}2$

Since, the slope of these equations are equal, the lines are parallel to each other. Proved

Given lines are:

3x + 5y = 7 i.e. 3x + 5y - 7 = 0........................(1)

3y = 2x + 4 i.e. 2x - 3y + 4 = 0........................(2)

Slope of eqn (1), m1 = - $\frac {x-coefficient}{y-coefficient}$ = $\frac {-2}{-3}$ = $\frac 23$

Slope of eqn (2), m2 = - $\frac {x-coefficient}{y-coefficient}$ = $\frac {-3}{5}$

Let $\theta$ be the angle between the equation (1) and (2):

tan$\theta$ =± $\frac {m_1 - m_2}{1 + m_1m_2}$

or, tan$\theta$ =± ($\frac {\frac 23 + \frac 35}{1 - \frac 23 × \frac 35}$)

or, tan$\theta$ =± ($\frac {\frac {10 + 9}{15}}{\frac {5 - 2}5}$)

or, tan$\theta$ =± ($\frac {19}{15}$ × $\frac 53$)

or, tan$\theta$ =± $\frac {19}9$

For acute angle,

tan$\theta$ = $\frac {19}9$= 2.11

∴ $\theta$ = 65°

∴ The acute angle between two lines is 65°. Ans

Here,

Given equation are:

3y - x - 6 = 0..............................(1)

y = 2x + 5 i.e. -2x + y = 5...................(2)

Slope of eqn (1), m1= -$\frac {x-coefficient}{y-coefficient}$ = -$\frac {(-1)}3$ = $\frac 13$

Slope of eqn (2), m2= -$\frac {x-coefficient}{y-coefficient}$ = -$\frac {(-2)}1$ = 2

If $\theta$ be the angle between the eqn (1) and (2),

tan$\theta$ =± $\frac {m_1 - m_2}{1 + m_1m_2}$

or, tan$\theta$ =± $\frac {\frac 13 - 2}{1 + \frac 13 × 2}$

or, tan$\theta$ =± $\frac {\frac {1 - 6}3}{\frac {3 + 2}3}$

or, tan$\theta$ =± $\frac {-5}3$× $\frac 35$

∴ tan$\theta$ =± (-1)

Taking -ve sign,

tan$\theta$ = +1

tan$\theta$ = 45°

∴$\theta$ = 45° Ans

Here,

x - 3y = 4......................(1)

2x - y = 3......................(2)

Slope of eqn (1), m1 = -$\frac {x-coefficient}{y-coefficient}$ = $\frac {-1}{-3}$ = $\frac 13$

Slope of eqn (2), m2 = -$\frac {x-coefficient}{y-coefficient}$ = $\frac {-2}{-1}$ = 2

If $\theta$ be the angle between the eqn (1) and (2),

tan$\theta$ =± $\frac {m_1 - m_2}{1 + m_1m_2}$

or, tan$\theta$ =± $\frac {\frac 13 - 2}{1 + \frac 13 × 2}$

or, tan$\theta$ =± $\frac {\frac {1 - 6}3}{\frac {3 + 2}3}$

or, tan$\theta$ =± $\frac {-5}3$× $\frac 35$

∴ tan$\theta$ =± (-1)

Taking -ve sign,

tan$\theta$ = +1

tan$\theta$ = 45°

∴$\theta$ = 45° Ans

Here,

Given equation are:

y - 3x - 2 = 0

or, -3x + y - 2 = 0..............................(1)

y = 2x + 5

or, - 2x + y = 5...................................(2)

Slope of eqn (1), m1 = -$\frac {x-coefficient}{y-coefficient}$ = $\frac {-3}{-1}$ = 3

Slope of eqn (2), m2 = -$\frac {x-coefficient}{y-coefficient}$ = -$\frac {-2}{1}$ =2

If $\theta$ be the angle between two lines,

tan$\theta$ =± $\frac {m_1 - m_2}{1 + m_1m_2}$

or, tan$\theta$ =± $\frac {3 - 2}{1 + 3 × 2}$

or, tan$\theta$ =± $\frac 1{1 + 6}$

∴ tan$\theta$ =± $\frac 17$

Taking +ve sign,

$\theta$ = tan-1($\frac 17$)

∴ $\theta$ = 8.13° Ans

Here,

Given lines are:

x = 3y + 8

i.e. x - 3y - 8 = 0..................................(1)

2x + 11 = 7y

i.e. 2x - 7y + 11 = 0............................(2)

Slope of eqn (1),m1= - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 1{-3}$ = $\frac 13$

Slope of eqn (2),m2= - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 2{-7}$ = $\frac 27$

Let $\theta$ be the angle between the equation (1) and (2),

tan$\theta$ =± $\frac {m_1 - m_2}{1 + m_1m_2}$

or, tan$\theta$ = ± $\frac {\frac 13 - \frac 27}{1 + \frac 13 × \frac 27}$

or, tan$\theta$ =± $\frac {\frac {7 - 6}{21}}{\frac {21 + 2}{21}}$

or, tan$\theta$ =± $\frac 1{21}$× $\frac {21}{23}$

∴ tan$\theta$ =± $\frac 1{23}$

For obtuse angle,

tan$\theta$ = -$\frac 1{23}$

or, tan$\theta$ = tan (180° -2°)

∴ $\theta$ = 178°

∴ The obtuse angle between two lines is 178°. Ans

Given equation is:

5x - 7y + 10 = 0............................(1)

Equation (1) changes in perpendicular form

7x + 5y + k = 0.............................(2)

The point (2, -1) passes through the equation (2)

7× 2 + 5× -1 + k = 0

or, 14 - 5 + k = 0

or, 9 + k = 0

∴ k = -9

Putting the value of k in equation (2)

7x + 5y - 9 = 0

∴ Required equation is:7x + 5y - 9 = 0 Ans

Given line is:

5x - 4y + 19 = 0...........................(1)

Equation (1) changes into the perpendicular form:

4x + 5y + k = 0............................(2)

The point (2, -3) passes through the equation (2)

4× 2 + 5× -3 + k = 0

or, 8 - 15 + k = 0

or, -7 + k = 0

∴ k = 7

Putting the value of k in equation (2)

4x + 5y + 7 = 0

∴ The required equation is 4x + 5y + 7 = 0. Ans

The slope of the line joining (3, - 7) and (-5, 3) is:

m1 = $\frac {y_2 - y_1}{x_2 - x_1}$ = $\frac {3 + 7}{-5 - 3}$ = $\frac {10}{-8}$ = -$\frac 54$

The mid-pointof the line joining (3, - 7) and (-5, 3) is:

P(x, y) = ($\frac {3 - 5}2$, $\frac {-7 + 3}2$) = ($\frac {-2}2$, $\frac {-4}2$) = (-1, -2)

The equation of the line passing through (-1, -2) is:

y + 2 = m (x + 1)............................(1)

Equation (1) is the perpendicular bisector of the line joining given two points,

m1× m2 = -1

or,-$\frac 54$× m2 = -1

∴ m2 = $\frac 45$

Putting the value of m2 in equation (1)

y + 2 = $\frac 45$(x + 1)

or, 5y + 10 = 4x + 4

or, 4x - 5y - 10 + 4 = 0

∴ 4x - 5y - 6 = 0 Ans

Here,

Given line is:

3x - 4y + 9 = 0..........................(1)

The line perpendicular to the equation (1) is:

4x + 3y + k = 0.........................(2)

The equation (2) passes through the point (-6, 4)

4× -6 + 3× 4 + k = 0

or, -24 + 12 + k = 0

or, -12 + k = 0

∴ k = 12

Putting the value of k in equation (2)

4x + 3y + 12 = 0 Ans

Given:

The co-ordinatesof A and B are: (3, -1) and (7, 1) respectively.

Equation of A(3, -1) and B(7, 1):

y + 1 = $\frac {1 + 1}{7 - 3}$ (x - 3) [$\because$ y - y1 = $\frac {y_2 - y_1}{x_2 - x_1}$ (x - x1)]

or, 4y + 4 = 2x - 6

or, 2x - 4y - 6 - 4 = 0

or, 2x - 4y - 10 = 0

or, 2(x - 2y - 5) = 0

∴ x - 2y - 5 = 0...................................(1)

Mid-point of A(3, -1) and B(7, 1)

= ($\frac {3 + 7}2$, $\frac {-1 + 1}2$) [$\because$ (x, y) = ($\frac {x_1 + x_2}2$, $\frac {y_1 + y_2}2$)]

= ($\frac {10}2$, $\frac 02$)

= (5, 0)

The perpendicular form of the equation x - 2y - 5 = 0 is:

2x + y + k = 0........................(2)

The point (5, 0) passes through the equation (2)

2x + y + k = 0

or, 2× 5 + 0 + k = 0

or, 10 + k = 0

∴ k = -10

Putting the value of k in equation (2),

2x + y - 10 = 0 Ans

Here,

Given equation is:

5x + 7y - 14 = 0.........................(1)

Equation (1) changes in perpendicular form

7x - 5y + k = 0............................(2)

Point (-2, -3) passes through the equation (2)

7× (-2) - 5× (-3) + k = 0

or, -14 + 15 + k = 0

or, 1 + k = 0

∴k = -1

Putting the value of k in equation (2)

7x - 5y - 1 = 0

∴ Required equation is:7x - 5y - 1 = 0 Ans

Given equation is:

4x - 3y = 10

i.e. 4x - 3y - 10 = 0..................................(1)

Equation (1) changes in the perpendicular form

3x + 4y + k = 0.........................................(2)

Point (2, 3) passes through the equation (2)

3× 2 + 4× 3 + k = 0

or, 6 + 12 + k = 0

or, 18 + k = 0

∴ k = -18

Substituting the value of k in equation (2)

3x + 4y - 18 = 0

∴ Required equation is: 3x + 4y - 18 = 0 Ans

Given equation is: 3x + 4y + 18 = 0.........................................(1)

Equation (1) changes in perpendicular form:

4x - 3y + k = 0.....................................(2)

Point (2, -3) passes through the equation (2)

4× 2 - 3× (-3) + k = 0

or, 8 + 9 + k = 0

or, 17 + k = 0

∴ k = -17

Putting the value of k in eqn (2)

4x - 3y - 17 = 0

∴ The required equation is:4x - 3y - 17 = 0 Ans

Given equation is: x - 2y - 2 = 0........................................(1)

Equation (1) changes in perpendicular form:

2x + y + k = 0.....................................(2)

Point (4, 6) passes through the equation (2)

2 × 4 + 6 + k = 0

or, 8 + 6+ k = 0

or, 14 + k = 0

∴ k = -14

Putting the value of k in eqn (2)

2x + y - 14 = 0

∴ The required equation is: 2x + y - 14 = 0Ans

Given equation is: x - 3y - 2 = 0............................(1)

Equation (1) changes in perpendicular form:

3x + y + k = 0.....................................(2)

Point (2, 3) passes through the equation (2)

3× 2+ 3+ k = 0

or, 6+ 3+ k = 0

or, 9+ k = 0

∴ k = -9

Putting the value of k in eqn (2)

3x + y -9 = 0

∴ The required equation is: 3x + y -9 = 0 Ans

Vertices of $\triangle$PQR are: P(-2, 1), Q(2, 3) and R(-2, -4).

We know that:

y - y1 = $\frac {y_2 - y_1}{x_2 - x_1}$(x- x1)

Eqn of QR,

y - 3 = $\frac {-4 - 3}{-2 - 2}$(x - 2)

or, y - 3 = $\frac {-7}{-4}$(x - 2)

or, y - 3 = $\frac {7}{4}$(x - 2)

or, 4y - 12 = 7x - 14

or, 7x - 14 - 4y +12 = 0

or, 7x - 4y - 2 = 0..................................(1)

The eqn (1) change in perpendicular form,

4x + 7y + k = 0.......................................(2)

The point (-2, 1) passes through eqn (1),

4× (-2) + 7× 1 + k = 0

or, -8 + 7 + k = 0

or, -1 + k = 0

∴ k = 1

Putting the value of k in eqn (2)

4x + 7y + 1 = 0 Ans

Given:

A(-1, 5), B(-4, -1) and C(3, -2) are the vertices of $\triangle$ABC.

We know,

Equation of the line passing through B(-4, -1) and C(3, -2) is:

y - y1 = $\frac {y_2 - y_1}{x_2 - x_1}$(x - x1)

or, y + 1 = $\frac {-2 + 1}{3 + 4}$(x + 4)

or, y + 1 = $\frac {-1}7$(x + 4)

or, 7 (y + 1) = -1 (x + 4)

or, 7y + 7 = -x - 4

or, x + 7y + 4 +7 = 0

∴ x + 7y + 11 = 0..................................(1)

The line parallel to the equation (1)

x + 7y + k = 0..........................................(2)

The equation (2) passes through the point A(-1, 5)

-1 + 7× 5 + k = 0

or, -1 + 35 + k = 0

or, 34 + k = 0

∴ k = -34

Putting the value of k in the equation (2)

x + 7y - 34 = 0

∴ The required equation is:x + 7y - 34 = 0 Ans

Here,

ABC is a triangle. The vertices of $\triangle$ABC are A(2, 8), B(-3, 5) and C(5, 3).

D is the mid-point of BC.

Co-ordinates of D

= ($\frac {x_1 + x_2}2$, $\frac {y_1 + y_2}2$)

= ($\frac {-3 + 5}2$, $\frac {5 + 3}2$)

= ($\frac 22$, $\frac 82$)

= (1, 4)

Slope of BC (m1)

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

= $\frac {3 - 5}{5 + 3}$

= $\frac {-2}8$

= $\frac {-1}4$

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

= $\frac {4 - 8}{1 - 2}$

= $\frac {-4}{-1}$

= 4

Now,

m1m2 = $\frac {-1}4$× 4 = -1

The product of two slope is equal to -1 so these are perpendicular.

Again,

y - y1 =$\frac {y_2 - y_1}{x_2 - x_1}$(x - x1)

or, y - 8 = $\frac {4 - 8}{1 - 2}$(x - 2)

or, y - 8 = $\frac {-4}{-1}$(x - 2)

or, y - 8 = 4x - 8

or, 4x - 8 - y + 8 = 0

∴ 4x - y = 0

Hence, the required equation is:4x - y = 0 Ans

Here,

The vertices of $\triangle$ABC are: A(3, 4), B(-2, 2) and C(3, -3).

Equation of BC is:

y - y1= $\frac {y_2 - y_1}{x_2 - x_1}$(x - x1)

or, y - 2 = $\frac {-3 - 2}{3 + 2}$(x + 2)

or, y - 2 = $\frac {-5}5$(x + 2)

or, y - 2 = -x - 2

or, x + 2 + y - 2 = 0

∴ x + y = 0.......................................(1)

The eqn (1) changes in perpendicular form,

x - y + k = 0......................................(2)

The point A(3, 4) passes through eqn (1)

3 - 4 + k = 0

or, -1 + k = 0

∴ k = 1

Putting the value of k in eqn (1)

x - y + 1 = 0

∴ The required equation is:x - y + 1 = 0 Ans

Given equation is: 7x - 24y + 10 = 0......................(1)

Equation (1) change in perpendicular form

24x + 7y + k = 0..............................(2)

Equation (2) passes through the point (-2, 4)

24× (-2) + 7× 4 + k = 0

or, -48 + 28 + k = 0

or, -20 + k = 0

∴ k = 20

Putting the value of k in equation (2)

24x + 7y + 20 = 0

Equation of the line PQ is: 24x + 7y + 20 = 0

We know,

d = $\begin {vmatrix} \frac {Ax + By + C}{\sqrt {A^2 + B^2}}\\ \end {vmatrix}$

Perpendicular length of PQ;

=$\begin {vmatrix} \frac {7 × (-2) - 24 × 4 + 10}{\sqrt {7^2 + (24)^2}}\\ \end {vmatrix}$

=$\begin {vmatrix} \frac {-14 - 96 + 10}{\sqrt {49 + 576}}\\ \end {vmatrix}$

=$\begin {vmatrix} \frac {-100}{\sqrt {625}}\\ \end {vmatrix}$

=$\begin {vmatrix} \frac {-100}{25}\\ \end {vmatrix}$

=$\begin {vmatrix} -4\\ \end {vmatrix}$

= 4 units

∴ The perpendicular length of PQ = 4 units and equation of PQ is:24x + 7y + 20 = 0 Ans

Let: m be the slope of the required line, so that its equation is:

y - y1 = m(x - x1).............................(1)

Point (2, -1) passes through equation (1):

y + 1 = m(x - 2)...............................(2)

Given equation is:

6x + 5y - 1 = 0.................................(3)

Slope of eqn (1) is:

slope (m1) = - $\frac {x-coefficient}{y-coefficient}$ = - $\frac 65$

Now,

Using angle formula,

tan$\theta$ =± ($\frac {m_1 - m_2}{1 + m_1m_2}$)

or, tan 45° =± ($\frac {m + \frac 65}{1 - m\frac 65}$)

or, 1 =± ($\frac {\frac {5m + 6}5}{\frac {5 - 6m}5}$)

or, 1 =± ($\frac {5m + 6}{5 - 6m}$)

Taking +ve sign,

5 - 6m = 5m + 6

or, -6m - 5m = 6 - 5

or, -11m = 1

∴ m = -$\frac 1{11}$

Taking -ve sign,

5 - 6m = - (5m + 6)

or, 5 - 6m = -5m - 6

or, -6m + 5m = - 6 - 5

or, - m = - 11

∴ m = 11

Putting the value of m = -$\frac 1{11}$ in equation (2)

y + 1 = m (x - 2)

or, y + 1 = -$\frac 1{11}$(x - 2)

or, 11y + 11 = -x + 2

or, x + 11y + 11 - 2 = 0

∴ x + 11y + 9 = 0

Putting the value of m = 11 in equation (2)

y + 1 = m(x - 2)

or, y + 1= 11 (x - 2)

or, y + 1 = 11x - 22

or, 11x - y - 22 - 1 = 0

∴ 1 1x - y - 23 = 0

Hence, the required equations are: x + 11y + 9 = 0 and 11x - y - 23 = 0 Ans

Given:

The eqn of AB is:

12(x + 3) = 5y

or, 12x + 36 - 5y = 0

or, 12x - 5y + 36 = 0........................(1)

Slope of eqn (1) is: m1 = $\frac {12}5$

The eqn of the line passes through (2, 3) is:

y - y1 = m (x - x1)

or, y - 3 = m (x - 2)..........................(2)

Slope of eqn (2) is: m2 = m

The angle between the lines (1) and (2) is 45°.

tan$\theta$ =± $\frac {m_1 - m_2}{1 + m_1m_2}$

or, tan 45° =± $\frac {\frac {15}2 - m}{1 + \frac {12}5m}$

or, 1 =± $\frac {\frac {12 - 5m}5}{\frac {5 + 12m}5}$

or, 1 =± $\frac {12 - 5m}{5 + 12m}$

Taking +ve sign,

5 + 12m = 12 - 5m

or, 12m + 5m = 12 - 5

or, 17m = 7

∴ m = $\frac 7{17}$

Taking -ve sign,

5 + 12m = - 12 + m

or, 12m - 5m = - 12 - 5

or, 7m = - 17

∴ m = -$\frac {17}7$

Substituting the value of m = $\frac 7{17}$ in eqn (2)

y - 3 = $\frac 7{17}$(x - 2)

or, 17(y - 3) = 7(x - 2)

or, 17y - 51 = 7x - 14

or, 7x - 17y + 51 - 14 = 0

∴ 7x - 17y + 37 = 0

Substituting the value of m = $\frac {-17}7$ in eqn (2)

y - 3 = $\frac {-17}7$(x - 2)

or, 7 (y - 3) = -17 (x - 2)

or, 7y - 21 = - 17x + 34

or, 17x - 34 + 7y - 21 = 0

∴ 17x + 7y - 55 = 0

∴ The required equations are:7x - 17y + 37 = 0 and17x + 7y - 55 = 0 Ans

Here,

The eqn of line is:

$\sqrt 3$x - y = 4.............................(1)

The slope of eqn (1) is: m1 =$\sqrt 3$

The eqn of the line passes through (4, 3) is:

y - 3 = m(x - 4)..........................(2)

Slope of eqn (2) is m2 = m

The angle between eqn (1) and (2) is 60°.

tan$\theta$ =± $\frac {m_1 - m_2}{1 + m_1m_2}$

or, tan 60° =± $\frac {\sqrt 3 -m}{1 + \sqrt3m}$

or, $\sqrt 3$ =± $\frac {\sqrt 3 - m}{1 + \sqrt 3m}$

Taking +ve sign,

$\sqrt 3$(1 + $\sqrt 3$m) = $\sqrt 3$ - m

or, $\sqrt 3$ + 3m = $\sqrt 3$ - m

or, 3m + m = $\sqrt 3$ - $\sqrt3$

or, 4m = 0

∴ m = 0

Taking -ve sign,

$\sqrt 3$ + 3m = -$\sqrt 3$ + m

or, 3m - m = -$\sqrt 3$ - $\sqrt 3$

or, 2m = - 2$\sqrt 3$

or, m = $\frac {-2\sqrt 3}2$

∴ m = -$\sqrt 3$

Putting the value of m = 0 in eqn (2)

y - 3 = m(x - 4)

or, y - 3 = 0(x - 4)

∴ y - 3 = 0

Putting the value of m = -$\sqrt 3$ in eqn (2)

y - 3 = m (x - 4)

or, y - 3 = -$\sqrt 3$ (x - 4)

or, y - 3 = -$\sqrt 3$x + 4$\sqrt 3$

∴ $\sqrt 3$x + y = 4$\sqrt 3$ + 3

∴ The required equations are: y - 3 = 0 and $\sqrt 3$x + y = 4$\sqrt 3$ + 3 Ans

Here,

The given eqn is:

2x + 3y + 5 = 0............................(1)

Slope of equation (1) is: m1 = $\frac {-2}3$

The equation of the line passes through (1, -4) is:

y + 4 = m(x - 1)...........................(2)

Slope of equation (2) is: m2 = m

The angle between the lines (1) and (2) is 45°.

tan$\theta$ =± ($\frac {m_1 - m_2}{1 + m_1m_2}$)

or, tan 45° =± ($\frac {\frac {-2}3 - m}{1 + m(\frac {-2}3)})$

or, 1 =± ($\frac {\frac {-2 - 3m}3}{\frac {3 - 2m}3})$

or, 1 =± ($\frac {-2 - 3m}{3 - 2m})$

Taking +ve sign,

1 = $\frac {-2 - 3m}{3 - 2m}$

or, -2 - 3m = 3 - 2m

or, -3m + 2m = 3 + 2

or, -m = 5

∴ m = -5

Taking -vesign,

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

or, 3 - 2m = 2 + 3m

or, 3m + 2m = 3 - 2

or, 5m = 1

∴ m = $\frac 15$

∴ m = -5, $\frac 15$

Substituting the value of m = -5 in equation (2)

y + 4 = - 5 (x - 1)

or, y + 4 = - 5x + 5

or, 5x + y + 4 - 5 = 0

∴ 5x + y - 1 = 0

Substituting the value of m = $\frac 15$ in equation (2)

y + 4 = $\frac 15$(x - 1)

or, 5y + 20 = x - 1

or, x - 1 - 5y - 20 = 0

∴ x - 5y - 21 = 0

Hence, the required equations are:5x + y - 1 = 0 andx - 5y - 21 = 0 Ans

Here,

The given equation is:

x - 3y = 2.............................(1)

The slope of eqn (1) is: m1 = -$\frac {x-coefficient}{y-coefficient}$ = $\frac {-1}{-3}$ = $\frac 13$

Let, the equation of the line passes through point (2, 3) is;

y - 3 = m(x - 2)....................(2) [$\because$y - y1 = m(x - x1)]

The slope of equation (2) is: m2 = m

If the angle between the lines (1) and (2) is 45°, then:

tan$\theta$ =± ($\frac {m_1 - m_2}{1 + m_1m_2})$

or, tan 45° =± ($\frac {\frac 13 - m}{1 + \frac 13m})$

or, 1 =± $\frac {\frac {1 - 3m}3}{\frac {3 + m}3}$

or, 1 =± $\frac {1 - 3m}{3 + m}$

Taking +ve sign,

1 = $\frac {1 - 3m}{3 + m}$

3 + m = 1 - 3m

or, 3m + m = 1 - 3

or, 4m = - 2

or, m = -$\frac 24$

∴ m = - $\frac 12$

Taking -ve sign,

1 = -$\frac {1 - 3m}{3 + m}$

or, 3 + m = - (1 - 3m)

or, 3 + m = -1 + 3m

or, 3m - m = 3 + 1

or, 2m = 4

or, m = $\frac 42$

∴ m = 2

Putting the value of m = - $\frac 12$ in eqn (2)

y - 3 = -$\frac 12$(x - 2)

or, 2y - 6 = - x + 2

or, x + 2y - 2 - 6 = 0

∴ x + 2y - 8 = 0

Putting the value of m = 2 in eqn (2)

y - 3 = 2(x - 2)

or, y - 3 = 2x - 4

or, 2x - 4 - y + 3 = 0

∴ 2x - y - 1 = 0

∴ The required equations are: x + 2y - 8 = 0 and2x - y - 1 = 0 Ans

Let: AB and AC be two lines whose equations are: y = m1x + c1 and y = m2x + c2, these lines meet OX in the point B and C respectively.

Let: $\angle$ABX = $\theta_1$ and $\angle$ = $\theta_2$

Let: tan$\theta_1$ = m1 and tan$\theta_2$ = m2

Let: $\angle$CAB = $\theta$, then;

$\theta_1$ = $\theta$ + $\theta_2$

∴ $\theta$ = $\theta_1$ - $\theta_2$

Taking tan on both sides,

tan$\theta$ = tan($\theta_1$ - $\theta_2$)

or, tan$\theta$ = $\frac {tan\theta_1 - tan\theta_2}{1 + tan\theta_1\theta_2}$

or, tan$\theta$ = $\frac {m_1 - m_2}{1 + m_1m_2}$

∴ $\theta$ = tan-1($\frac {m_1 - m_2}{1 + m_1m_2}$) Ans

Given equations are:

a1x + b1y + c1 = 0............................(1)

a2x + b2y + c2 = 0............................(2)

Let: m1 and m2 represents the slope of the lines (1) and (2) respectively.

Slope of (1), m1= -$\frac {x-coefficient}{y-coefficient}$ = -$\frac {a_1}{b_1}$

Slope of (2), m2= -$\frac {x-coefficient}{y-coefficient}$ = -$\frac {a_2}{b_2}$

Let: $\theta$ be the angle between the two given lines. Then,

tan$\theta$ =± ($\frac {m_1 - m_2}{1 + m_1m_2})$

or, tan$\theta$ =± ($\frac {(\frac {-a_1}{b_1}) - (\frac {-a_2}{b_2})}{1 + (\frac {-a_1}{b_1})(\frac {-a_2}{b_2})})$

or, tan$\theta$ =± ($\frac {\frac {-a_1b_2 + a_2b_1}{b_1b_2}}{\frac {b_1b_2 + a_1a_2}{b_1b_2}})$

or, tan$\theta$ =± ($\frac {a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2})$

∴ $\theta$ = tan-1± ($\frac {a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2})$ Ans

Let: ABC is an equilateral triangle with base BC on x-axis.

i.e. y = 0, $\angle$CBA = 60° and $\angle$XCA = 120°

Slope of line AB = m1 = tan$\theta$ = tan 60° = $\sqrt 3$

Slope of line CA = m2 = tan$\theta$ = tan 120° = - $\sqrt 3$

The eqn of AB is:

y - y1 = m(x - x1)

or, y - 2 = $\sqrt 3$(x + 1)

or, y - 2 = $\sqrt 3$x + $\sqrt 3$

∴$\sqrt 3$x - y + 2 + $\sqrt 3$ = 0

The eqn of the sides CA is:

y - y1 = m(x - x1)

or, y - 2 = -$\sqrt 3$ (x + 1)

or, y - 2 = -$\sqrt 3$x - $\sqrt 3$

∴$\sqrt 3$x + y - 2 + $\sqrt 3$ = 0

Hence, the required lines are:$\sqrt 3$x - y + 2 + $\sqrt 3$ = 0 and$\sqrt 3$x + y - 2 + $\sqrt 3$ = 0 Ans

Given equations are:

x + 2y = 3...........................(1)

2x - 3y = 20......................(2)

Eqn (1) is multipliedby 2 and subtract with eqn (2)

 2x + 4y = 6 2x - 3y = 20 - + - 7y = -14

y = -$\frac {14}7$

∴ y = -2

Putting the value of y in eqn (1)

x + 2y = 3

or, x + 2× (-2) = 3

or, x - 4 = 3

or, x = 3 + 4

∴ x = 7

The point of intersection is: (7, -2)

The given eqn is:

2x - 3y + 5 = 0.............................(3)

The eqn (3) changes in perpendicular form:

-3x - 2y + k = 0..........................(4)

The point (7, -2) passes through eqn (4)

-3× 7 - 2× (-2) + k = 0

or, -21 + 4 + k = 0

or, -17 + k = 0

∴ k = 17

Putting the value of k in eqn (4)

-3x - 2y + 17 = 0

∴ 3x + 2y - 17 = 0 Ans

Given points are: (-3, -4) and (7, 1)

Eqn of two points is:

y - y1 = $\frac {y_2 - y_1}{x_2 - x_1}$(x - x1)

or, y + 4 = $\frac {1 + 4}{7 + 3}$(x + 3)

or, y +4 = $\frac 5{10}$(x + 3)

or, y +4 = $\frac 12$(x + 3)

or, 2y + 8 = x + 3

or, x + 3 - 2y - 8 = 0

∴ x - 2y - 5 = 0.............................(1)

The points (-3, -4) and (7, 1) divides by a point P(x, y) in the ratio 3 : 2.

Using section formula,

P(x, y) = ($\frac {m_1x_2 + m_2x_1}{m_1 + m_2}$, $\frac {m_1y_2 + m_2y_1}{m_1 + m_2})$

or,P(x, y) = ($\frac {3 × 7 + 2 × (-3)}{3 + 2}$, $\frac {3 × 1 + 2 × (-4)}{3 + 2}$)

or,P(x, y) = ($\frac {21 - 6}5$, $\frac {3 - 8}5$)

or,P(x, y) = ($\frac {15}5$, $\frac {-5}5$)

∴ P(x, y) = (3, -1)

The eqn (1) changes in perpendicular form is:

-2x - y + k = 0....................................(2)

The point (3, -1) passes through eqn (2)

-2× 3 - (-1) + k = 0

or, -6 + 1 + k = 0

or, -5 + k = 0

∴ k = 5

Putting the value of k in eqn (2)

-2x - y + 5 = 0

∴ 2x + y - 5 = 0 Ans

Here,

The diagonals of the square bisect each other.

Mid-point of AC = ($\frac {x_1 + x_2}2$, $\frac {y_1 + y_2}2$) = ($\frac {2 - 6}2$, $\frac {3 + 5}2$) = ($\frac {-4}2$, $\frac 82$) = (-2, 4)

The eqn of the points (2, 3) and (-6, 5) is:

y - y1 = $\frac {y_2 - y_1}{x_2 - x_1}$(x - x1)

or, y - 3 = $\frac {5 - 3}{-6 - 2}$(x - 2)

or, y - 3 = $\frac 2{-8}$(x - 2)

or, y - 3 = $\frac 1{-4}$(x - 2)

or, - 4y + 12 = x - 2

or, x - 2 + 4y - 12 = 0

∴ x + 4y - 14 = 0.............................(1)

Since, the diagonals of the square are perpendicular to each other.

The eqn (1) change in perpendicular form:

4x - y + k = 0.....................................(2)

The point (-2, 4) passes through eqn (2)

4× (-2) - 4 + k = 0

or, -8 - 4 + k = 0

or, - 12 + k = 0

∴ k = 12

Putting the value of k in eqn (2)

4x - y + 12 = 0

∴ The required equation is:4x - y + 12 = 0 Ans