Notes on Angle between two lines | Grade 10 > Optional Mathematics > Co-ordinate Geometry | KULLABS.COM

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#### 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.

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

.

### Very Short Questions

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

0%

180(^0)

135 (^0)

145(^0)

140(^0)

• ### What will be  the formula of the angle between the lines y = m1x+C1 and y = m2x + c2.

θ = (tan-^1) (frac{m1+m2}{±1+m1+m2})

θ = (tan-^1) (frac{m2-m1}{±1-m1+m2})

θ = (tan-^1) (frac{m2-m1}{±1-m1-m2})

θ = (tan-^1) (frac{m1-m2}{±1+m1-m2})

• ### If the straight lines px+3y-12 = 0 and 4y-3x+7 = 0 are parallel to each other , find the value of P.

(frac{9}{4})

(frac{3x}{4y})

(frac{-9}{4})

(frac{-9}{-4})

3

4

6

2

• ### What will be the slope of the straight line perpedicular to 4x+3y = 12.

(frac{4}{3})

(frac{2}{3})

(frac{6}{12})

(frac{3}{4})

• ### If the straight lines 2x+3y+6 = 0 and ax-5y+20 = 0 are perpendicular to each other , find the value of a .

(frac{15}{3})

(frac{3}{4})

(frac{15}{2})

(frac{2}{17})

5x+7y+31 = 0

5y+7x+13 = 0

5x+31+y7 = 0

5y+7x+13 = 0

9x+ y= 4

4x+ y= 9

9x+ 9= y

9x+ y= 9

3x+4y = 18

4x+3y = 81

18x+4y = 0

4x+ 3y= 9

• ### Find the equation of straight lines passing through a point (-6 , 4) and perpendicular to the line 3x-4y + 9 = 0 .

4y+3x+0 = 12

3x+4y+12 = 12

4x+3y+12 = 0

0+3y+12 = 4x

5x-7y = 44

44x-5y = 7x

7x-5y = 44

7x+5y = 44

3y+13+9x = 0

3y-13-0 = 9x

9x+3y+13 = 0

13y+9x+3 = 0

• ### Find the equation of the straight lines passing through the point (2 , 3) and making an angle of 45(^0) with the line x-3y = 2.

2x-1y = 1 , 8-2y = 0

1-2xy = 0 , 2y-8 = x

2x-y = 1 , x+2y=8

1-2xy = 0 , 2y-8 = x

• ### Find the equation of the straight lines passing through the point (1,0) & inclined at an angle of 30(^0) with the line x- (sqrt{3y}) = 4.

y=0 and (sqrt{4x})-y = (sqrt{4})

y=0 and (sqrt{1x})-y = (sqrt{1})

y=0 and (sqrt{-3x})-y = (sqrt{-3})

y=0 and (sqrt{3x})-y = (sqrt{3})

1, -1

-1, -1

0, 0

1, 1

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##### Find the equation of the straight line passing through the point (5,0)and making 45° with the line 4x-5y 9=0

Find the equation of the straight line passing through the point(5,0)and making 45°withthe line 4x-5y 9=0