Angle between two lines

Angle between the lines y = m₁ + c₁ and y = m_{2 +} 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 =θ₁-θ₂

∴ tanΦ = tan(θ₁ -θ₂) = \(\frac {tan\theta_1 - tan\theta_2}{1 + tan\theta_1 tan\theta_2}\) = \(\frac{m_1 - m_2}{1+tan\theta_1 tan\theta_2}\) .........(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 = 90^o.

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 + B_{1y +} 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 + B_{1y +} 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.