Subject: Optional Mathematics

**The angle between two lines**

**Angle between the lines y = m _{1 }+ c_{1} and y = m_{2 + }c_{2} and y = m_{2}x + c_{2}**

**Condition of perpendicularity**

**m _{1}m_{2} = -1**

**Condition of Parallelism**

m_{1 }= m_{2}

Angle between the lines A_{1}x + B_{1}y + C_{1} = 0 and A_{2}x + B_{2}y +C_{2} = 0

**Condition of perpendicularity**

A_{1}A_{2} + B_{1}B_{2} = 0

**Condition of Parallelism**

A_{1}B_{2 }= A_{2}B_{1}

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

k = -bc

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

k = ac

Let the equation of two lines AB and CD be y = m_{1}x + c_{1 }and y = m_{2}x + c_{2} respectively.

let the lines AB and CD make angles θ_{1} and θ_{2} respectively with the positive direction of X-axis.

Then, tanθ_{1} = m_{1} and tanθ_{2} = m_{2}.

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, θ_{1} = Φ + θ_{2}

or,Φ =θ_{1}-θ_{2}

∴ tanΦ = tan(θ_{1 }-θ_{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 = 180^{0}

or, Ψ = 180^{0} - 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 = m_{1}x + c_{1} and y = m_{2}x + c_{2} 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, tan90^{0} = ± \(\frac{m_1 - m_2}{1+m_1m_2}\)

or, cot90^{0} = ± \(\frac{1+m_1m_2}{m_1 - m_2}\)

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

or, 1 +m_{1}m_{2} =0

or, m_{1}m_{2} = -1

Two lines will be perpendicular to each other if m_{1}m_{2} = -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 = 0^{0}.

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

or, tan0^{0} = ± \(\frac{m_1 - m_2}{1+m_1m_2}\)

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

or, m_{1} - m_{2} = 0

or, m_{1 }= m_{2}

∴ Two lines will be parallel to each other if m_{1} = m_{2} i.e. if slopes are equal.

Let equations of two straight lines AB and CD be A_{1}x + B_{1y + }C_{1} = 0 and A_{2}x + B_{2}y + C_{2} = 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 θ_{1} and θ_{2} with the positive direction of X-axis.

Then, tanθ_{1} = -\(\frac{A_1}{B_1}\) and tanθ_{2} = -\(\frac{A_2}{B_2}\)

Let ∠CEA = Φ.

Then, θ_{1} = θ_{1} - θ_{2}

or, Φ = θ_{1} - θ_{2}

∴ tan Φ = tan( θ_{1} - θ_{2}) = \(\frac{tan\theta_1 - tan\theta_2}{1 + tan\theta_1 tan\theta_2}\) = \(\frac{A_2 B_1 - A_1 B_2}{A_1 A_2 + B_1 B_2}\) = -\(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\) ......(i)

Let ∠BEC = Ψ

Then Ψ + Φ = 180^{0}

or, Ψ = 180^{0} - Φ

∴ tan Ψ = tan(180^{0} - 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 A_{1}x + B_{1y + }C_{1} = 0 and A_{2}x + B_{2}y + C_{2} = 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 = 90^{0}

Then tan90^{0} = ± \(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\)

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

∴ A_{1}A_{2} + B_{1}B_{2} = 0

**Condition of Parallelism**

Two lines AB and CD will be parrallel to each other if θ = 0^{0}

Then, tan0^{0} = ± \(\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}\)

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

or, A_{1}B_{2} - A_{2}B_{1} = 0

or, A_{1}B_{2 }= A_{2}B_{1}

∴ \(\frac{A_1}{A_2}\) = \(\frac{B_1}{B_2}\)

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 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 = m _{1 }+ c_{1} and y = m_{2 + }c_{2} and y = m_{2}x + c_{2}**

**Condition of perpendicularity**

m_{1}m_{2} = -1

**Condition of Parallelism**

m_{1 }= m_{2}

Angle between the lines A_{1}x + B_{1}y + C_{1} = 0 and A_{2}x + B_{2}y +C_{2} = 0

**Condition of perpendicularity**

A_{1}A_{2} + B_{1}B_{2} = 0

**Condition of Parallelism**

A_{1}B_{2 }= A_{2}B_{1}

**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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Write down the condition for the lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 to the perpendicular and parallel to each other.

Given:

a_{1}x + b_{1}y + c_{1} = 0..............................(1)

a_{2}x + b_{2}y + c_{2} = 0..............................(2)

Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {a_1}{b_1}\)

Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {a_2}{b_2}\)

when the lines are parallel, then:

m_{1} = m_{2}

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

∴ a_{1}b_{2} = a_{2}b_{1 Ans}

when the lines are perpendicular, then:

m_{1}× m_{2}= - 1

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

∴ a_{1}a_{2} = - b_{1}b_{2}_{Ans}

Prove that the two straight lines 5x + 4y - 10 = 0 and 15x + 12y - 7 = 0 are parallel to each other.

Here,

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

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

Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 54\)

Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {15}{12}\) = -\(\frac 54\)

m_{1} = m_{2} = - \(\frac 54\)

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

Prove that the pair of straight lines x + 3y = 2 and 6x - 2y = 9 are perpendicular to each other.

Here,

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

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

Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 13\)

Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 6{-2}\) =3

We have,

m_{1}× m_{2} =- \(\frac 13\)× 3 = - 1

∴m_{1}× m_{2} = - 1

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

If two lines 3x - 2y - 5 = 0 and 2x + py - 3 = 0 are parallel to each other, find the value pf p.

Here,

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

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

Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 3{-2}\) = \(\frac 32\)

Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2{p}\)

When two lines are parallel lines, then:

m_{1} = m_{2}

or,\(\frac 32\) = - \(\frac 2{p}\)

or, p = \(\frac {-2 × 2}3\)

∴ p = -\(\frac 43\) _{Ans}

If two lines 4x + ky - 4 = 0 and 2x - 6y = 5 are perpendicular to each others find the value of k.

Here,

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

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

Slope of eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 4k\)

Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2{-6}\) = \(\frac 13\)

when two line are perpendicular to each other,

m_{1}× m_{2} = - 1

or, - \(\frac 4k\) × \(\frac 13\) = - 1

or, - \(\frac 43\)× - 1 = k

∴ k = \(\frac 43\) _{Ans}

Write down the formula for the angle between the pair of lines y = m_{1}x + c_{1} and y = m_{2}x + c_{2}, stating also the condition when they are perpendicular and parallel to each other.

The formulae of angle between y = m_{1}x + c_{1} andy = m_{2}x + c_{2}is:

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

when m_{1}× m_{2} = -1, the two lines are perpendicular to each other.

when m_{1} = m_{2}, the two lines are parallel to each other.

If the line passing through (3, -4) and (-2, a) is parallel to the line given by the equation y + 2x + 3 = 0. Find the value of a.

Here,

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

m_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {a + 4}{-2 - 3}\) = \(\frac {-(a + 4)}5\)

Given eq^{n} is y + 2x + 3 = 0

Slope of above eq^{n} (m_{2}) = - \(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 21\) = - 2

when lines are parallel then,

m_{1} = m_{2}

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

or, a + 4 = 10

or, a = 10 - 4

∴ a = 6 _{Ans}

Prove that the line joining the points (3, -4) and (-2, 6) is parallel to the line y + 2x + 3 = 0.

Here,

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

m_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {6 + 4}{-2 - 3}\) = \(\frac {10}{-5}\) = -2

Slope of the eq^{n} y + 2x + 3 = 0 is:

m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 21\) = -2

From above,

m_{1}= m_{2} = - 2

Hence, the lines are parallel. _{Proved}

For what the value of k, the line kx - 3y + 6 = 0 is perpendicular to the lie joining (4, 3) and (5, -3)?

Here,

Given eq^{n} is kx - 3y + 6 = 0

Slope of above eq^{n} is:

m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac k{-3}\) = \(\frac k3\)

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

m_{2} = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {-3 - 3}{5 - 4}\) = - \(\frac 61\) = -6

If lines are perpendicular then:

m_{1}× m_{2} = -1

or, \(\frac k3\)× -6 = -1

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

∴ k = \(\frac 12\) _{Ans}

Write down the formula for the angle between the pair of lines y = m_{1}x + c_{1} and y = m_{2}x + c_{2}, stating also the condition when they are perpendicular to each other.

Here,

Given equations of the lines are:

y = m_{1}x + c_{1}...............................(1)

y = m_{2}x + c_{2}...............................(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 + m_{1}m_{2} = 0

∴ m_{1}m_{2} = -1 _{Ans}

If two lines 2x + ay + 3 = 0 and 3x - 2y = 5 are perpendicular to each other find the value of a.

Here,

The given equations are:

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

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

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

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

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

m_{1}× m_{2} = -1

or,- \(\frac 2a\)×\(\frac 32\) = -1

or, -6 = - 2a

or, a = \(\frac 62\)

∴ a = 3 _{Ans}

Prove that 2x + 4y - 7 = 0 and 6x + 12y + 4 = 0 are parallel to each other.

Here,

Given equations of the lines are:

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

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

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

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

∴ m_{1} = m_{2} = \(\frac {-1}2\)

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

Find the actual angle between the lines 3x + 5y = 7 and 3y = 2x + 4.

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 eq^{n} (1), m_{1} = - \(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-2}{-3}\) = \(\frac 23\)

Slope of eq^{n} (2), m_{2} = - \(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-3}{5}\)

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}

Find the acute angle between lines 3y - x - 6 = 0 and y = 2x + 5.

Here,

Given equation are:

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

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

Slope of eq^{n} (1), m_{1}= -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {(-1)}3\) = \(\frac 13\)

Slope of eq^{n} (2), m_{2}= -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {(-2)}1\) = 2

If \(\theta\) be the angle between the eq^{n} (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}

Find the acute angle between the lines x - 3y = 4 and 2x - y = 3.

Here,

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

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

Slope of eq^{n} (1), m_{1} = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-1}{-3}\) = \(\frac 13\)

Slope of eq^{n} (2), m_{2} = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-2}{-1}\) = 2

If \(\theta\) be the angle between the eq^{n} (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}

In the given figure, find the value of \(\theta\).

Here,

Given equation are:

y - 3x - 2 = 0

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

y = 2x + 5

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

Slope of eq^{n} (1), m_{1} = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-3}{-1}\) = 3

Slope of eq^{n} (2), m_{2} = -\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac {-2}{1}\) =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}

Find the obtuse angle between the lines x = 3y + 8 and 2x + 11 = 7y.

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 eq^{n }(1),^{}m_{1}= - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 1{-3}\) = \(\frac 13\)

Slope of eq^{n }(2),^{}m_{2}= - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 2{-7}\) = \(\frac 27\)

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}

Find the equation of the lines passing the point (2, -1) and perpendicular to the line 5x - 7y + 10 = 0.

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}

Find the equation of the line passing through the point (2, -3) and perpendicular to the line 5x - 4y + 19 = 0.

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}

Find the equation of the perpendicular bisector of the line joining point (3, -7) and (-5, 3).

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

m_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\) = \(\frac {3 + 7}{-5 - 3}\) = \(\frac {10}{-8}\) = -\(\frac 54\)

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,

m_{1}× m_{2} = -1

or,-\(\frac 54\)× m_{2} = -1

∴ m_{2} = \(\frac 45\)

Putting the value of m_{2} 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}

Find the equation of the straight line, which passes through the point (-6, 4) and perpendicular to 3x - 4y + 9 = 0.

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}

The points A and B have co-ordinates (3, -1) and (7, 1) respectively. Find the equation of the perpendicular bisector of AB.

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 - y_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\) (x - x_{1})]

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}

Find the equation of the straight line which passes through the point (-2, -3) and perpendicular to the line 5x + 7y = 14.

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}

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

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}

Find the equation of the line passing through the point (2, -3) and perpendicular to the line 3x + 4y + 18 = 0.

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 eq^{n} (2)

4x - 3y - 17 = 0

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

Find the equation of the line which passes through the point (4, 6) and is perpendicular to the line x - 2y - 2 = 0.

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 eq^{n} (2)

2x + y - 14 = 0

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

Find the equation of a straight line passing through the point (2, 3) and perpendicular to the straight line x - 3y - 2 = 0.

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 eq^{n} (2)

3x + y -9 = 0

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

\(\triangle\)PQR has its vertices P(-2, 1), Q(2, 3) and R(-2, -4). Find the equation of the perpendicular drawn from the vertex P to QR.

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

We know that:

y - y_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\)(x- x_{1})

Eq^{n} 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 eq^{n} (1) change in perpendicular form,

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

The point (-2, 1) passes through eq^{n} (1),

4× (-2) + 7× 1 + k = 0

or, -8 + 7 + k = 0

or, -1 + k = 0

∴ k = 1

Putting the value of k in eq^{n} (2)

4x + 7y + 1 = 0 _{Ans}

A(-1, 5), B(-4, -1) and C(3, -2) are the vertices of \(\triangle\)ABC. Find the equation of the line which passes through the vertex A and parallel to the side BC.

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 - y_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x_{1})

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}

In the given figure D is the mid-point of BC. Prove that: AD ⊥ BC. Find the equation of AD also.

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 (m_{1})

= \(\frac {y_2 - y_1}{x_2 - x_1}\)

= \(\frac {3 - 5}{5 + 3}\)

= \(\frac {-2}8\)

= \(\frac {-1}4\)

Slope of AD (m_{2})

= \(\frac {y_2 - y_1}{x_2 - x_1}\)

= \(\frac {4 - 8}{1 - 2}\)

= \(\frac {-4}{-1}\)

= 4

Now,

m_{1}m_{2} = \(\frac {-1}4\)× 4 = -1

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

∴ AD⊥ BC _{Proved}

Again,

Equation of AD;

y - y_{1} =\(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x_{1})

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}

The vertices of A(3, 4), B(-2, 2) and C(3, -3) of a \(\triangle\)ABC. AD is perpendicular drawn from the vertex A on the opposite side BC. Find the equation of the AD.

Here,

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

Equation of BC is:

y - y_{1}= \(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x_{1})

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 eq^{n} (1) changes in perpendicular form,

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

The point A(3, 4) passes through eq^{n} (1)

3 - 4 + k = 0

or, -1 + k = 0

∴ k = 1

Putting the value of k in eq^{n} (1)

x - y + 1 = 0

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

From the point P(-2, 4), if PQ is drawn perpendicular to the line 7x - 24y + 10 = 0. Find the equation of the line PQ. Also determine the length of PQ.

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}

Find the equation of two lines which passes through (2, -1) and make an angle of 45° with the lines 6x + 5y - 1 = 0.

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

y - y_{1} = m(x - x_{1}).............................(1)

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

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

Given equation is:

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

Slope of eq^{n} (1) is:

slope (m_{1}) = - \(\frac {x-coefficient}{y-coefficient}\) = - \(\frac 65\)

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}

In the given figure the equation of AB is 12(x + 3) = 5y, find the equation of AC.

Given:

The eq^{n} of AB is:

12(x + 3) = 5y

or, 12x + 36 - 5y = 0

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

Slope of eq^{n} (1) is: m_{1} = \(\frac {12}5\)

The eq^{n} of the line passes through (2, 3) is:

y - y_{1} = m (x - x_{1})

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

Slope of eq^{n} (2) is: m_{2} = 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 eq^{n} (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 eq^{n} (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}

Find the equation of the straight lines passing through the point (4, 3) and making an angle of 60° to the line \(\sqrt 3\)x - y = 4.

Here,

The eq^{n} of line is:

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

The slope of eq^{n} (1) is: m_{1} =\(\sqrt 3\)

The eq^{n} of the line passes through (4, 3) is:

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

Slope of eq^{n} (2) is m_{2} = m

The angle between eq^{n} (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 eq^{n} (2)

y - 3 = m(x - 4)

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

∴ y - 3 = 0

Putting the value of m = -\(\sqrt 3\) in eq^{n} (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}

Find the equation of straight line which passes through the point (1, -4) and make an angle of 45° with the straight line 2x + 3y + 5 = 0.

Here,

The given eq^{n} is:

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

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

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

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

Slope of equation (2) is: m_{2} = 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}

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

Here,

The given equation is:

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

The slope of eq^{n} (1) is: m_{1} = -\(\frac {x-coefficient}{y-coefficient}\) = \(\frac {-1}{-3}\) = \(\frac 13\)

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

y - 3 = m(x - 2)....................(2) [\(\because\)y - y_{1} = m(x - x_{1})]

The slope of equation (2) is: m_{2} = 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 eq^{n} (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 eq^{n} (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}

Find the angle between two lines whose equations are y = m_{1}x + c_{1} and y = m_{2}x + c_{2}.

Let: AB and AC be two lines whose equations are: y = m_{1}x + c_{1} and y = m_{2}x + c_{2}, these lines meet OX in the point B and C respectively.

Let: \(\angle\)ABX = \(\theta_1\) and \(\angle\) = \(\theta_2\)

Let: tan\(\theta_1\) = m_{1} and tan\(\theta_2\) = m_{2}

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}

Find the angle between two lines whose equation are a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0.

Given equations are:

a_{1}x + b_{1}y + c_{1} = 0............................(1)

a_{2}x + b_{2}y + c_{2} = 0............................(2)

Let: m_{1} and m_{2} represents the slope of the lines (1) and (2) respectively.

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

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

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}

Find the equation of the sides of an equilateral triangle whose vertex is (-1, 2) and base is y = 0.

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 = m_{1} = tan\(\theta\) = tan 60° = \(\sqrt 3\)

Slope of line CA = m_{2} = tan\(\theta\) = tan 120° = - \(\sqrt 3\)

The eq^{n} of AB is:

y - y_{1} = m(x - x_{1})

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

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

∴\(\sqrt 3\)x - y + 2 + \(\sqrt 3\) = 0

The eq^{n} of the sides CA is:

y - y_{1} = m(x - x_{1})

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}

Find the equation of the straight line passing through the point of intersection of the lines x + 2y = 3 and 2x - 3y = 20 and perpendicular to the line 2x - 3y + 5 = 0.

Given equations are:

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

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

Eq^{n} (1) is multipliedby 2 and subtract with eq^{n} (2)

2x | + | 4y | = | 6 |

2x | - | 3y | = | 20 |

- | + | - | ||

7y | = | -14 |

y = -\(\frac {14}7\)

∴ y = -2

Putting the value of y in eq^{n} (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 eq^{n} is:

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

The eq^{n} (3) changes in perpendicular form:

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

The point (7, -2) passes through eq^{n} (4)

-3× 7 - 2× (-2) + k = 0

or, -21 + 4 + k = 0

or, -17 + k = 0

∴ k = 17

Putting the value of k in eq^{n} (4)

-3x - 2y + 17 = 0

∴ 3x + 2y - 17 = 0 _{Ans}

Find the equation of the straight line passing through the point that divides the join of (-3, 4) and (7, 1) in the ratio 3 : 2 and perpendicular to the line.

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

Eq^{n} of two points is:

y - y_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x_{1})

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 eq^{n} (1) changes in perpendicular form is:

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

The point (3, -1) passes through eq^{n} (2)

-2× 3 - (-1) + k = 0

or, -6 + 1 + k = 0

or, -5 + k = 0

∴ k = 5

Putting the value of k in eq^{n} (2)

-2x - y + 5 = 0

∴ 2x + y - 5 = 0 _{Ans}

If (2, 3) and (-6, 5) are the ends points of the diagonal of a square. Find the equation of the other diagonal.

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 eq^{n} of the points (2, 3) and (-6, 5) is:

y - y_{1} = \(\frac {y_2 - y_1}{x_2 - x_1}\)(x - x_{1})

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 eq^{n} (1) change in perpendicular form:

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

The point (-2, 4) passes through eq^{n} (2)

4× (-2) - 4 + k = 0

or, -8 - 4 + k = 0

or, - 12 + k = 0

∴ k = 12

Putting the value of k in eq^{n} (2)

4x - y + 12 = 0

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

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