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

Angle between the lines
Angle between the lines

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

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

fdgs

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

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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


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lines 2x-3ay=4 and x 2y=1 are perpendicular to each other.

Lines 2x-3ay=4 and x 2y=1 are perpendicular to each other.


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