Straight Lines

The equation of line in intercept form is:

where a and b are the intercepts on the axis.

Given that: a = b

⇒ x + y = a …(i)

If eq. (i) passes through the point (1, - 2), we get

1 + (-2) = a

⇒ 1 – 2 = a

⇒ a = -1

Putting the value of ‘a’ in eq. (i), we get

x + y = -1

⇒ x + y + 1 = 0

Hence, the equation of straight line is x + y + 1 = 0 which passes through the point (1, – 2).

Given points are A(5, 2), B(2,3) and C(3, -1)

Firstly, we find the slope of the line joining the points (2,3) and (3, -1)

It is given that line passing through the point (5,2) is perpendicular

to BC

∵ m₁m₂ = -1

⇒ -4 × m₂ = -1

Now, we have to find the equation of line passing through point (5,2)

Equation of line: y – y₁ = m(x – x₁)

⇒ 4y – 8 = x – 5

⇒ x – 5 – 4y + 8 = 0

⇒ x – 4y + 3 = 0

Hence, the equation of line passing through the point (5,2) is x – 4y + 3 = 0

Given equations are:-

y = (2 - √3)(x + 5)

⇒ y = (2 - √3)x + (2 - √3)5 …(i)

and y = (2 + √3)(x – 7)

⇒ y = (2 + √3)x – 7(2 + √3) …(ii)

Now, we have to find the slope of eq. (i)

Since, the eq. (i) is in y = mx + b form, we can easily see that the slope (m₁) is (2 - √3)

Now, the slope (m₂) of eq. (ii) is (2 + √3)

Let θ be the angle between the given two lines.

Putting the values of m₁ and m₂ in above eq., we get

[∵(a – b)(a + b) = (a² – b²)]

⇒ tan θ = √3

⇒ θ = tan^-1(√3)

⇒ θ = 60°

Hence, the required angle is 60°

The equation of line in intercept form is:

…(i)

where a and b are the intercepts on the axis.

Given that: a + b = 14

⇒ b = 14 – a

So, equation of line is

⇒ 14x – ax + ay = 14a – a² …(ii)

If eq. (ii) passes through the point (3,4) then

14(3) – a(3) + a(4) = 14a – a²

⇒ 42 – 3a + 4a – 14a + a² = 0

⇒ a² – 13a + 42 = 0

⇒ a² – 7a – 6a + 42 = 0

⇒ a(a – 7) – 6(a – 7) = 0

⇒ (a – 6)(a – 7) = 0

⇒ a – 6 = 0 or a – 7 = 0

⇒ a = 6 or a = 7

If a = 6, then

6 + b = 14

⇒b = 14 – 6 = 8

If a = 7, then

7 + b = 14

⇒b = 14 – 7

= 7

If a = 6 and b = 8, then equation of line is

⇒ 4x + 3y = 24

If a = 7 and b = 7, then equation of line is

⇒ x + y = 7

Let (x₁,y₁) be any point lying in the equation x+ y = 4

∴ x₁ + y₁ = 4 …(i)

Distance of the point (x₁,y₁) from the equation 4x + 3y = 10

[given]

⇒ 4x₁ + 3y₁ – 10 = ±5

either 4x₁ + 3y₁ – 10 = 5 or 4x₁ + 3y₁ – 10 = -5

4x₁ + 3y₁ = 5 + 10 or 4x₁ + 3y₁ = -5 + 10

4x₁ + 3y₁ = 15 …(ii) or 4x₁ + 3y₁ = 5 …(iii)

From eq. (i), we have y₁ = 4 – x₁ …(iv)

Putting the value of y₁ in eq. (ii), we get

4x₁ + 3(4 – x₁) = 15

⇒ 4x₁ + 12 – 3x₁ = 15

⇒ x₁ = 15 – 12

⇒ x₁ = 3

Putting the value of x₁ in eq. (iv), we get

y₁ = 4 – 3

⇒ y₁ = 1

Putting the value of y₁ = 4 – x₁ in eq. (iii), we get

4x₁ + 3(4 – x₁) = 5

⇒ 4x₁ + 12 – 3x₁ = 5

⇒ x₁ = 5 – 12

⇒ x₁ = - 7

Putting the value of x₁ in eq. (iv), we get

y₁ = 4 – (-7)

⇒ y₁ = 4 + 7

⇒ y₁ = 11

Hence, the required points on the given line are (3,1) and (-7,11)

Given: …(i)

and …(ii)

Firstly, we find the slope of the given lines

Since, the above equation is in y = mx + b form.

So, Slope of the eq. (i) is

Now, finding the slope of the eq. (ii)

Since, the above equation is in y = mx + b form.

So, Slope of the eq. (ii) is

Let θ be the angle between the given two lines.

Putting the values of m₁ and m₂ in above eq., we get

Hence, the required angle is

Hence Proved

Given that line passing through (1, 2) making an angle 30° with y – axis.

∴ Angle made by the line with x – axis is (90° - 30°) = 60°

∴ Slope of the line, m = tan 60°

= √3

So, the equation of the line passing through the point (x₁, y₁) and having slope ‘m’ is

y – y₁ = m(x – x₁)

Here, (x₁, y₁) = (1, 2) and m = √3

⇒ y – 2 = √3(x – 1)

⇒ y – 2 = √3x - √3

⇒ √3x – y - √3 + 2 = 0

Given lines are:

2x + y = 5 …(i)

x + 3y = -8 …(ii)

Firstly, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (ii) by 2, we get

2x + 6y = -16 …(iii)

On subtracting eq. (iii) from (i), we get

2x + y – 2x – 6y = 5 – (-16)

⇒ -5y = 5 + 16

⇒ -5y = 21

Putting the value of y in eq. (i), we get

⇒ 10x = 46

Hence, the point of intersection is

Now, we find the slope of the given equation 3x + 4y = 7

We know that the slope of an equation is

So, the slope of a line which is parallel to this line is also

Then the equation of the line passing through the point having slope is:

y – y₁ = m (x – x₁)

⇒ 3x + 4y + 3 = 0

Given equation is ax + by + 8 = 0

It can also be re-written as ax + by = -8

Now, dividing by -8 to both the sides, we get

So, the intercepts on the axes are

Now, the second equation which is given is 2x – 3y + 6 = 0

It can also be re-written as 2x – 3y = -6

Now, dividing by -6 to both the sides, we get

So, the intercepts are -3 and 2

Now, according to the question

[intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes]

Therefore, and

and

Let a and b be the intercepts on the given line

∴ Coordinates of A and B are (a, 0) and (0, b) respectively.

Given that coordinate axes is divided by the point (-5, 4) in the ratio 1:2.

Now, using the section formula, we find the value of a and b

⇒-15 = 2a and b = 12

∴ Coordinates of A and B are

Now, we have to find the equation of line AB.

Equation of line when two points are given:

Putting the values, we get

⇒ 5y = 8x + 60

⇒ 8x – 5y + 60 = 0

Hence, the required equation is 8x – 5y + 60 = 0

Given: Length of the perpendicular from the origin (OM) = 4 units

and line makes an angle with positive direction of x – axis

∠BAX = 120°

∴ ∠BAO = 180° - 120° = 60°

∴ ∠MAO = 60°

Now, In Δ AMO,

∠MAO + ∠AOM + ∠OMA = 180°

[∵ sum of angles of a triangle is 180°]

⇒ 60° + θ + 90° = 180°

⇒ 150° + θ = 180°

⇒ θ = 180° - 150°

⇒ θ = 30°

∴ ∠AOM = 30°

Now, we find the equation in NORMAL FORM

xcosθ + ysinθ = p

⇒ xcos(30°) + ysin(30°) = 4

[given: p = 4]

⇒ √3x + y = 8

Hence, the required equation is √3x + y = 8

Given that equation of the hypotenuse is 3x + 4y = 4

and opposite vertex of the hypotenuse is (2, 2)

Firstly, we find the slope of the given equation

3x + 4y = 4

It can be re-written as 4y = 4 – 3x

Since, the above equation is in y = mx + b form

So,

Now, let the slope of AC be m

Now, we find the value of m, by using the formula

Putting the values of m₁ and m₂ in above eq., we get

[∵ tan 45° = 1]

Taking (+) sign, we get

⇒ 4m + 3 = 4 – 3m

⇒ 4m + 3m = 4 – 3

⇒ 7m = 1

Taking (-) sign, we get

⇒ 4m + 3 = - (4 – 3m)

⇒ 4m + 3 = - 4 + 3m

⇒ 4m – 3m = - 4 – 3

⇒ m = -7

If , then equation of AC is

y – y₁ = m(x – x₁)

⇒ 7y – 14 = x – 2

⇒ x – 7y – 2 + 14 = 0

⇒ x – 7y + 12 = 0

If m = -7, then equation of AC is

y – 2 = (-7)(x – 2)

⇒ y – 2 = -7x + 14

⇒ 7x + y = 16

Hence, the required equations are x – 7y + 12 = 0 and 7x + y = 16

Let Δ ABC be an equilateral triangle.

Given: Equation of the base BC is x + y = 2

We know that, in an equilateral triangle all angles are of 60°

So, in Δ ABD

We know that,

the distance d of a point P(x₀, y₀) from the line Ax + By + C = 0 is given by

Now, length of perpendicular from vertex A(2, -1) to the line x + y = 2 is

Squaring both the sides, we get

Hence, the required length of side is

Let the variable line be ax + by = 1

We know that, length of the perpendicular from (p, q) to the line ax + by + c = 0 is

Now, perpendicular distance from A(2, 0)

Now, perpendicular distance from B(0, 2)

Now, perpendicular distance from C (1, 1)

It is given that the algebraic sum of the perpendicular from the given points (2, 0), (0, 2) and (1, 1) to this line is zero.

d₁ + d₂ + d₃ = 0

⇒ 2a – 1 + 2b – 1 + a + b – 1 = 0

⇒ 3a + 3b – 3 = 0

⇒ a + b – 1 = 0

⇒ a + b = 1

So, the equation ax + by = 1 represents a family of straight lines passing through a fixed point.

Comparing the equation ax + by = 1 and a + b = 1, we get

x = 1 and y = 1

So, the coordinates of fixed point is (1, 1)

Let the given line x + y = 4 and the required line ‘l’ intersect at B(a, b)

Slope of line ‘l’ is

and we also know that, m = tanθ

…(i)

Given that:

So, by distance formula for point A(1, 2) and B(a, b), we get

On squaring both the sides, we get

⇒ 2 = 3a² + 3 – 6a + 3b² + 12 – 12b

⇒ 2 = 3a² + 3b² – 6a – 12b + 15

⇒ 3a² + 3b² – 6a – 12b + 13 = 0 …(ii)

Point B(a, b) also satisfies the eq. x + y = 4

∴ a + b = 4

⇒ b = 4 – a …(iii)

Putting the value of b in eq. (ii), we get

3a² + 3(4 – a)² – 6a – 12(4 – a) + 13 = 0

⇒ 3a² + 3(16 + a² – 8a) – 6a – 48 + 12a + 13 = 0

⇒ 3a² + 48 + 3a² – 24a – 6a – 48 + 12a + 13 = 0

⇒ 6a² – 18a + 13 = 0

Now, we solve the above equation by using this formula

Putting the value of a in eq. (iii), we get

Now, putting the value of a and b in eq. (i), we get

We solve the eq. (A) to get the value of θ, we get

We know that,

We have,

θ = tan^-1(√3) - tan^-1(1)

θ = tan^-1(tan 60°) – tan^-1(tan 45°)

θ = 60° - 45°

θ = 15°

Now, we solve the eq. B.

We know that,

We have,

θ = tan^-1(√3) + tan^-1(1)

θ = tan^-1(tan 60°) + tan^-1(tan 45°)

θ = 60° + 45°

θ = 105°

Intercepts form of a straight line is

where a and b are the intercepts on the axes

Given that:

this shows that the line is passing through the fixed point (k, k)

Let AB be a line passing through a point (-4, 3) and meets x – axis at A (a, 0) and y – axis at B (0, b)

Using the section formula for internal division, i.e.

…(i)

Here, m₁ = 5, m₂ = 3

(x₁, y₁) = (a, 0) and (x₂, y₂) = (0, b)

Putting the above values in the above formula, we get

⇒ -32 = 3a or 24 = 5b

or

Intercept form of the line is

Putting the value of a and b in above equation, we get

⇒ -72x + 160y = 768

⇒ -36x + 80y = 384

⇒ 18x – 40y + 192 = 0

⇒ 9x – 20y + 96 = 0

Hence, the required equation is 9x – 20y + 96 = 0

Given two lines are:

x – y + 1 = 0 …(i)

and 2x – 3y + 5 = 0 …(ii)

Now, point of intersection of these lines can be find out as:

Multiplying eq. (i) by 2, we get

2x – 2y + 2 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

2x – 2y + 2 – 2x + 3y – 5 = 0

⇒ y – 3 = 0

⇒ y = 3

On putting value of y in eq. (ii), we get

2x – 3(3) + 5 = 0

⇒ 2x – 9 + 5 = 0

⇒ 2x – 4 = 0

⇒ 2x = 4

⇒ x = 2

So, the point of intersection of given two lines is:

(x, y) = (2, 3)

Let m be the slope of the required line

∴ Equation of the line is

y – 3 = m(x – 2)

⇒ y – 3 = mx – 2m

⇒ mx – y – 2m + 3 = 0 …(i)

Since, the perpendicular distance from the point (3, 2) to the line is then

Squaring both the sides, we get

⇒ 49(m² + 1) = 25(m + 1)²

⇒ 49m² + 49 = 25(m² + 1 + 2m)

⇒ 49m² + 49 = 25m² + 25 + 50m

⇒ 25m² + 25 + 50m – 49m² – 49 = 0

⇒ – 24m² + 50m – 24 = 0

⇒ – 12m² + 25m – 12 = 0

⇒ 12m² – 25m + 12 = 0

⇒ 12m² – 16m – 9m + 12 = 0

⇒ 4m (3m – 4) – 3(3m – 4) = 0

⇒ (3m – 4)(4m – 3) = 0

⇒ 3m – 4 = 0 or 4m – 3 = 0

⇒ 3m = 4 or 4m = 3

or

Putting the value of in eq. (i), we get

⇒ 4x – 3y + 1 = 0

Putting the value of in eq. (i), we get

⇒ 3x – 4y + 6 = 0

hence, the required equation are 4x – 3y + 1 = 0 and 3x – 4y + 6 = 0

Let the coordinates of a moving point P be (a, b)

Given that the sum of the distance from the axes to the point is always 1

∴ |x| + |y| = 1

⇒ ±x ± y = 1

⇒ – x – y = 1, x + y = 1, -x + y = 1 and x – y = 1

Hence, these equations gives us the locus of the point P which is a square.

Given lines are y - √3|x| = 2

If x ≥ 0, then

y - √3x = 2 …(i)

If x < 0, then

y + √3x = 2 …(ii)

On adding eq. (i) and (ii), we get

y - √3x + y + √3x = 2 + 2

⇒ 2y = 4

⇒ y = 2

Putting the value of y = 2 in eq. (ii), we get

2 + √3x = 2

⇒ √3x = 2 – 2

⇒ x = 0

∴ Point of intersection of given lines is (0, 2)

Now, we find the slopes of given lines.

Slope of eq. (i) is

y = √3x + 2

Comparing the above equation with y = mx + b, we get

m = √3

and we know that, m = tan θ

∴ tan θ = √3

⇒ θ = 60° [∵ tan 60° = √3]

Slope of eq. (ii) is

y = - √3x + 2

Comparing the above equation with y = mx + b, we get

m = -√3

and we know that, m = tan θ

∴ tan θ = -√3

⇒ θ = (180° - 60°)

⇒ θ = 120°

In ΔACB,

[given: AC = 5units]

∴ OB = OA + AB

Hence, the coordinates of the foot of perpendicular =

Given equation is

Since, p is the length of perpendicular drawn from the origin to the given line

Squaring both the sides, we have

…(i)

Since, a², b² and p² are in AP

∴ 2p² = a² + b²

…(ii)

Form eq. (i) and (ii), we get

⇒ (a² + b²)(a² + b²) = 2(a²b²)

⇒ a⁴ + b⁴ + a²b² + a²b² = 2a²b²

⇒ a⁴ + b⁴ = 0

Hence Proved

Given that:

We know that,

Slope of a line, m = tan θ

Since, the lines cut off intercepts – 3 on y – axis then the line is passing through the point (0, -3).

So, the equation of line is

y – y₁ = m(x – x₁)

⇒ 5y + 15 = 3x

⇒ 5y – 3x + 15 = 0

Hence, the correct option is (a)

The equation of line in intercept form is:

where a and b are the intercepts on the axis.

Given that: a = b

⇒ x + y = a

⇒ y = - x + a

⇒ y = (-1)x + a

Since, the above equation is in y = mx + b form

So, the slope of the line is – 1.

Given that straight line passing through the point (3, 2)

and perpendicular to the line y = x

Let the equation of line ‘L’ is

y – y₁ = m(x – x₁)

Since, L is passing through the point (3, 2)

∴ y – 2 = m(x – 3) …(i)

Now, given eq. is y = x

Since, the above equation is in y = mx + b form

So, the slope of this equation is 1

It is also given that line L and y = x are perpendicular to each other.

We know that, when two lines are perpendicular, then

m₁ × m₂ = -1

∴ m × 1 = -1

⇒ m = -1

Putting the value of m in eq. (i), we get

y – 2 = (-1)(x – 3)

⇒ y – 2 = -x + 3

⇒ x + y = 3 + 2

⇒ x + y = 5

Hence, the correct option is (b)

Given that line passing through the point (1, 2)

and perpendicular to the line x + y + 1 = 0

Let the equation of line ‘L’ is

x – y + k = 0 …(i)

Since, L is passing through the point (1, 2)

∴ 1 – 2 + k = 0

⇒ k = 1

Putting the value of k in eq. (i), we get

x – y + 1 = 0

or y – x – 1 = 0

Hence, the correct option is (b)

Let the first equation of line having intercepts on the axes a, -b is

⇒ bx – ay = ab …(i)

Let the second equation of line having intercepts on the axes b, -a is

⇒ ax – by = ab …(ii)

Now, we find the slope of eq. (i)

bx – ay = ab

⇒ ay = bx – ab

Since, the above equation is in y = mx + b form

So, the slope of eq. (i) is

Now, we find the slope of eq. (ii)

ax – by = ab

⇒ by = ax – ab

Since, the above equation is in y = mx + b form

So, the slope of eq. (i) is

Let θ be the angle between the given two lines.

Putting the values of m₁ and m₂ in above eq., we get

Hence, the required angle is

Hence, the correct option is (c)

Given points are (2, -3) and (4, -5)

Firstly, we find the equation of line.

We know that,

Equation of line when two points are given:

Putting the values, we get

⇒y + 3 = - 1 (x – 2)

⇒ y + 3 = - x + 2

⇒ x + y = 2 – 3

⇒ x + y = - 1

(Intercept form)

Comparing the above equation with the given equation , we get the value of a and b

a = -1 and b = -1

Hence, the correct option is (d)

Given two lines are:

2x – 3y + 5 = 0 …(i)

and 3x + 4y = 0 …(ii)

Now, point of intersection of these lines can be find out as:

Multiplying eq. (i) by 3, we get

6x – 9y + 15 = 0 …(iii)

Multiplying eq. (ii) by 2, we get

6x + 8y = 0 …(iv)

On subtracting eq. (iv) from (iii), we get

6x – 9y + 15 – 6x – 8y = 0

⇒ – 17y + 15 = 0

⇒ - 17y = -15

On putting value of y in eq. (ii), we get

So, the point of intersection of given two lines is:

Now, perpendicular distance from the point to the given line 5x – 2y = 0

Hence, the correct option is (a)

Given equation is √3x + y = 1

and θ = 60°

Firstly, we find the slope of the given equation

√3x + y = 1

⇒ y = 1 - √3x

or y = (-√3)x + 1

Since, the above equation is in y = mx + b form.

So, the slope of this equation m₁ = -√3

Let m₂ be the slope of the required line.

Now, we find the value of m₂, by using the formula

Putting the values of m₁ and m₂ in above eq., we get

[∵ tan 60° = √3]

Taking (+) sign, we get

⇒ -√3 – m₂ = √3(1 - √3m₂)

⇒ -√3 – m₂ = √3 - 3m₂

⇒ 3m₂ – m₂ = √3 + √3

⇒ 2m₂ = 2√3

⇒ m₂ = √3

Taking (-) sign, we get

⇒ √3 + m₂ = √3(1 - √3m₂)

⇒ √3 + m₂ = √3 - 3m₂

⇒ 3m₂ + m₂ = 0

⇒ 4m₂ = 0

⇒ m₂ = 0

∴, Equation of line passing through (3, -2) with slope √3 is

y – y₁ = m(x – x₁)

⇒ y + 2 = √3x – 3√3

⇒ √3x – y – 3√3 – 2 = 0

⇒ √3x – y – (3√3 + 2) = 0

and Equation of line passing through (3, -2) with slope 0 is

y – y₁ = m(x – x₁)

⇒y – (-2) = 0 (x – 3)

⇒ y + 2 = 0

Hence, the required equations are √3x – y – (3√3 + 2) = 0 and y + 2 = 0

Hence, the correct option is (a)

Let equation of any line passing through the point (1, 0) is

y – y₁ = m(x – x₁)

⇒ y – 0 = m(x – 1)

⇒ y = mx – m

⇒ mx – y – m = 0 …(i)

Given that distance of the line from origin is

Squaring both sides, we get

⇒ 3(m² + 1) = 4m²

⇒ 3m² + 3 = 4m²

⇒ 4m² – 3m² = 3

⇒ m² = 3

⇒ m = ±√3

Putting the value of m = √3 in eq. (i), we get

√3x – y – √3 = 0

Now, putting the value of m = -√3 in eq. (i), we get

-√3x – y – (-√3) = 0

⇒ -√3x – y + √3 = 0

Hence, the correct option is (a)

Given equations are y = mx + c₁ …(i)

and y = mx + c₂ …(ii)

Firstly, we find the slope of eq. (i) and (ii)

Since, both the equations have same slope i.e. m

So, they are parallel lines.

We know that, distance between two parallel lines

Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is

Hence, the correct option is (b)

Given equation is y = 3x + 4 …(i)

Since, this equation is in y = mx + b form.

So, slope (m₁) of the given equation is 3

Let equation of any line passing through the point (2, 3) is

y – y₁ = m(x – x₁)

⇒ y – 3 = m(x – 2) …(ii)

Given that eq. (i) is perpendicular to eq. (ii)

And we know that, if two lines are perpendicular then,

m₁m₂ = -1

⇒ 3 × m₂ = -1

Putting the value of slope in eq. (ii), we get

⇒ 3y – 9 = -x + 2

⇒ x + 3y – 9 – 2 = 0

⇒ x + 3y – 11 = 0 …(iii)

Now, we have to find the coordinates of foot of the perpendicular.

Solving eq. (i) and (iii), we get

x + 3(3x + 4) – 11 = 0 [from(i)]

⇒ x + 9x + 12 – 11 = 0

⇒ 10x + 1 = 0

Putting the value of x in eq. (i), we get

So, the required coordinates are

Hence, the correct option is (b)

Let the given line meets the axes at A (a, 0) and B (0, b)

Given that (3, 2) is the midpoint

⇒ a = 6

and

⇒ b = 4

Intercept form of the line AB is

Putting the value of a and b in above equation, we get

⇒ 2x + 3y = 12

Hence, the correct option is (a)

Given equation of line is y = 3x – 1

Now, we find the slope of the above equation.

Since, the above equation is in y = mx + b form

So, m = 3

Now, we find the equation of line passing through the point (1, 2) and parallel to the given line with slope = 3

y – y₁ = m(x – x₁)

⇒ y – 2 = 3(x – 1)

Hence, the correct option is (c)

Given lines are x = 0, y = 0, x = 1 and y = 1 form a square of side 1 unit

By the given lines we form a square OABC having corners O (0, 0),

A (1, 0), B (1, 1) and C (0, 1).

Now, we find the equation of diagonal AC

We know that, when two points are given then, equation of line is

⇒ y = -1 (x – 1)

⇒ y = -x + 1

⇒ x + y = 1

Equation of diagonal OB is

⇒ y = 1 (x)

⇒ y = x

Hence, the correct option is (a)

Different form of equation of straight line are

Intercept form:

where, a and b are the intercepts on the axis

In intercept form, we need two parameters ‘a’ and ‘b’ to specify a straight line

Slope – Intercept Form:

y = mx + c

where m = tan θ, and θ is the angle made with positive x – axis

and c is the intercept on y – axis.

So, we need two parameters ‘m’ and ‘c’ to specify a straight line.

Hence, the correct option is (b)

Let Q(x, y) be the reflection of P(4, 1) about the line y = x, then midpoint of PQ

which lies on y = x

⇒ 4 + x = 1 + y

⇒ x – y + 3 = 0 …(i)

Now, we find the slope of given equation y = x

Since, this equation is in y = mx + b form.

So, the slope = m = 1

Slope of PQ =

Since, PQ is perpendicular to y = x

And we know that, when two lines are perpendicular then

m₁ m₂ = -1

⇒ y – 1 = - (x – 4)

⇒ y – 1 = - x + 4

⇒ x + y – 5 = 0 …(ii)

On adding eq. (i) and (ii), we get

x – y + 3 + x + y – 5 = 0

⇒ 2x – 2 = 0

⇒ x – 1 = 0

⇒ x = 1

Putting the value of x = 1 in eq. (i), we get

1 – y + 3 = 0

⇒ -y + 4 = 0

⇒ y = 4

Given that translation through a distance 2 units along the positive x-axis

∴ The point after translation is (1 + 2, 4) = (3, 4)

Hence, the correct option is (b)

Given equations are:

4x + 3y + 10 = 0 …(i)

5x – 12y + 26 = 0 …(ii)

and 7x + 24y – 50 = 0 …(iii)

Let (p, q) be the point which is equidistant from the given lines.

Now, we find the distance of (p, q) from the given lines.

We know that,

the distance d of a point P(x₀, y₀) from the line Ax + By + C = 0 is given by

Distance of (p, q) from eq. (i) is

Distance of (p, q) from eq. (ii) is

Distance of (p, q) from eq. (iii) is

Given that (p, q) is equidistant from the given lines

Let put the value of (p, q) = (0, 0) in above equation, we get

Hence, the correct option is (c)

Given line is 3x + y = 3

Now, we find the slope of given equation 3x + y = 3

It can be re-written as y = -3x + 3

Since, the above equation is in y = mx + b form

So, the slope m = -3

So, the line passes through (2, 2) and having slope is

⇒ 3y – 6 = x – 2

⇒ 3y = x + 4

[∵y = mx + c]

So, the y – intercept is

Hence, the correct option is (d)

Given lines are:

3x + 4y + 5 = 0 …(i)

3x + 4y – 5 = 0 …(ii)

and 3x + 4y + 2 = 0 …(iii)

Clearly, eq. (i), (ii) and (iii) are parallel to each other as the coefficients of x and y are same.

We know that,

distance between two parallel lines is

Distance between parallel lines

3x + 4y + 5 = 0

3x + 4y + 2 = 0

Distance between parallel lines

3x + 4y – 5 = 0

3x + 4y + 2 = 0

∴ Ratio between the distance is

Hence, the correct option is (b)

Let ABC be an equilateral triangle with vertex A (a, b)

Let AD ⊥ BC and let (p, q) be the coordinates of D

Given that the centroid P lies at the origin (0, 0).

We know that, the centroid of a triangle divides the median in the ratio 1: 2

Now, using the section formula, we get

⇒ a + 2p = 0 and b + 2q = 0 …(A)

⇒ a + 2p = b + 2q

⇒ 2p – 2q = b – a

⇒ 2(p – q) = b – a …(i)

It is given that BC = x + y – 2 = 0

Since, the above equation passes through (p, q)

⇒ p + q – 2 = 0 …(ii)

Now, we find the slope of line AP

and equation of line BC is

x + y – 2 = 0

⇒ y = - x + 2

⇒ y = (-1)x + 2

Since the above equation is in y = mx + b form.

So, slope of line BC is

Since, both the lines are perpendicular to each other.

∴ m_AP × m_BC = -1

⇒ b = a

Putting the value of b = a in eq. (i), we get

2(p – q) = b – b

⇒ 2(p – q) = 0

⇒ p = q

Now, putting the value p = q in eq. (ii), we get

p + p – 2 = 0

⇒ 2p = 2

⇒ p = 1

⇒ q = 1 [∵ p = q]

Putting the value of p and q in eq. (A), we get

a + 2 × 1 = 0 and b + 2 × 1 = 0

⇒ a = -2 and ⇒ b = -2

So, the coordinates of vertex A (a, b) is (-2, -2)

Hence, the correct option is (c)

Given: a, b, c are in AP

⇒ 2b = a + c

⇒ a – 2b + c = 0 …(i)

Now, comparing the eq. (i) with the given equation ax + by + c = 0, we get

x = 1, y = -2

So, the line will pass through (1, -2)

Ans. If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through (1, -2).

The equation of line in intercept form is:

where a and b are the intercepts on the axis.

Given that: a = b

⇒ x + y = a …(i)

If eq. (i) passes through the point (1, - 2), we get

1 + (-2) = a

⇒ 1 – 2 = a

⇒ a = -1

Putting the value of ‘a’ in eq. (i), we get

x + y = -1

⇒ x + y + 1 = 0

Ans. The line which cuts off equal intercept from the axes and pass through the point (1, –2) is x + y + 1.

Given equation is:-

x – 2y = 3

⇒ x – 3 = 2y

…(i)

Now, we have to find the slope of eq. (i)

Since, the eq. (i) is in y = mx + b form.

So, slope of eq. (i) is

Now, we have to find the equation which is passing through the point (3, 2).

We know that, if a line passing through the point (x₁, y₁) then the equation of line is

y – y₁ = m (x – x₁)

So, here x₁ = 3 and y₁ = 2

∴ y – 2 = m(x – 3) …(ii)

Now, it is given that the angle between the given two lines is 45°.

Putting the values of m₁ and m₂ in above eq., we get

⇒ 2m – 1 = 2 + m or -(2m – 1) = 2 + m

⇒ 2m – m = 2 + 1 or -2m + 1 – m = 2

⇒ m = 3 or -3m = 1

or

Putting the value of m = 3 in eq. (ii), we get

y – 2 = 3(x – 3)

⇒ y – 2 = 3x – 9

⇒ 3x – y – 9 + 2 = 0

⇒ 3x – y – 7 = 0

Putting the value of in eq. (ii), we get

⇒ 3(y – 2) = 3 – x

⇒ 3y – 6 = 3 – x

⇒ x + 3y – 6 – 3 = 0

⇒ x + 3y – 9 = 0

Given line: 3x – 4y – 8 = 0

and given points are (3, 4) and (2, -6)

For point (3, 4)

3(3) – 4(4) – 8

= 9 – 16 – 8

= 9 – 24

= - 15 < 0

For point (2, -6)

3(2) – 4(-6) – 8

= 6 + 24 – 8

= 30 – 8

= 22 > 0

Ans. So, the points (3,4) and (2, -6) are situated on the opposite sides of 3x – 4y – 8 = 0.

Given point is (3, -2) and equation of line is 5x – 12y = 3

Let (p, q) be any moving point.

∴ Distance between (p, q) and (3, -2)

⇒ (d₁)² = (p – 3)² + (q + 2)²

Now, distance of the point (p, q) from the given line 5x – 12y – 3 = 0 is

According to the question, we have (d₁)² = d₂

Taking numerical values only, we have

⇒ 13[(p – 3)² + (q + 2)²] = 5p – 12q – 3

⇒ 13[p² + 9 – 6p + q² + 4 + 4q] = 5p – 12q – 3

⇒ 13p² + 117 – 78p + 13q² + 52 + 52q = 5p – 12q – 3

⇒ 13p² + 13q² – 78p + 52q + 169 = 5p – 12q – 3

⇒ 13p² + 13q² – 78p + 52q + 169 – 5p + 12q + 3 = 0

⇒ 13p² + 13q² – 83p + 64q + 172 = 0

Ans. A point moves so that square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is 13p² + 13q² – 83p + 64q + 172 = 0

Given equation of the line is

x sin θ + y cos θ = p …(i)

Let P(h, k) be the midpoint of the given line where it meets the two axis at (a, 0) and (0, b).

Since, (a, 0) lies on eq. (i) then

a sin θ + 0 = p

…(ii)

(0, b) also lies on the eq. (i) then

0 + b cos θ = p

…(iii)

Since, P(h, k) is the midpoint of the given line

⇒ 2h = a

and

⇒ 2k = b

Putting the value of a = 2h in eq. (ii), we get

…(iv)

Putting the value of b = 2k in eq. (ii), we get

…(v)

Squaring and adding eq. (iv) and (v), we get

[∵ sin²θ + cos²θ = 1]

or

or 4x²y² = p²y² + p²x²

or 4x²y² = p²(x² + y²)

Ans. Locus of the mid-points of the portion of the line x sin θ + y cos θ = p intercepted between the axes is 4x²y² = p²(x² + y²)

Let ABC be a triangle with vertices A(x_1, y₁), B (x_2, y₂) and C (x_2, y₂), where x_i, y_i, i = 1, 2, 3 are integers

Then, Area of ΔABC

Since, x_i and y_i all are integers but is a rational number. So, the result comes out to be a rational number.

i.e. Area of ΔABC = a rational number

Suppose, ABC be an equilateral triangle, then Area of ΔABC is

[∵ AB = BC = CA]

It is given that vertices are integral coordinates, it means the value of coordinates is in whole number. Therefore, the value of (AB)² is also an integer.

But, √3 is an irrational number.

⇒ Area of ΔABC = an ir-rational number

This is a contradiction to the fact that the area is a rational number.

Hence, the given statement is TRUE

Given points are A (– 2, 1), B (0, 5) and C (– 1, 2)

To prove: A (– 2, 1), B (0, 5) and C (– 1, 2) are collinear

Proof: Two ways to find that given points are collinear or not:

I^st way: If three points are collinear, then slope of any two pairs of points will be equal.

II^nd way: If the value of area of triangle formed by the three points is zero, then the points are collinear.

So, we check by the 1^st method

Slope of AB

Slope of BC

Slope of CA

Since, the slopes are different. So, the given points are not collinear.

Hence, the given statement is FALSE

Let the equation of line y = mx + c …(i)

So, slope of the above equation is ‘m’

Given equation of line is x sec θ + y cosec θ = a

⇒ y cosecθ = a – x secθ

Since the above equation is in y = mx + b form

So, slope of the equation is

Given that eq. (i) is perpendicular to x sec θ + y cosec θ = a

⇒ m × m’ = -1

Putting the value of m in eq. (i), we get

⇒ y secθ = x cosecθ + c (secθ)

⇒ x cosecθ – y secθ = - c secθ

⇒ x cosecθ – y secθ = k …(ii)

[Let k = - c secθ]

If eq. (ii) passes through the point (a cos³θ, a sin³θ)

⇒ (a cos³θ) cosecθ – (a sin³θ) secθ = k

⇒ a[(cos²θ – sin²θ)(cos²θ + sin²θ)] = xcosθ – ysinθ

[∵(a⁴ – b⁴) = (a² – b²)(a² + b²)]

⇒ a[(cos²θ – sin²θ)(1)] = xcosθ – ysinθ

[∵cos²θ + sin²θ = 1]

⇒ a[cos 2θ] = xcosθ – ysinθ

[∵cos²θ – sin²θ = cos 2θ]

⇒ xcosθ – ysinθ = a cos 2θ

Hence, the given statement is FALSE

Given equation are

x + 2y – 10 = 0 …(i)

and 2x + y + 5 = 0 …(ii)

Firstly, we find the point of intersection:

Multiplying the eq. (i) by 2, we get

2x + 4y – 20 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

2x + y + 5 – 2x – 4y + 20 = 0

⇒ -3y + 25 = 0

⇒ -3y = -25

Putting the value of y in eq. (i), we get

⇒ 3x + 20 = 0

If the given line 5x + 4y = 0 passes through the point then

⇒ 0 = 0

So, the given line passes through the point of intersection of the given lines.

Hence, the given statement is TRUE

Let ABC be an equilateral triangle with vertex (2, 3)

and equation of opposite side is x + y = 2

We know that, all angle of an equilateral triangle is of 60°

So, θ = 60°

Let the slope of line AB is m

and slope of the given equation x + y = 2 is

∴ m₂ = -1

We know that,

Putting the values of m₁ and m₂ in above eq., we get

⇒ (1 – m)√3 = 1 + m or (1 – m)(-√3) = – 1 – m

⇒ √3 - √3 m = 1 + m or -√3 + √3 m = – 1 – m

⇒ √3 – 1 = m + √3m or -√3 + 1 = - m - √3m

⇒ √3 – 1 = m(1 + √3) or -(√3 – 1) = -m(√3 + 1)

…(i) or …(ii)

Now, multiply and divide eq. (i) by √3 + 1, we get

⇒ m = 2 + √3

Now, multiply and divide eq. (i) by √3 – 1 , we get

⇒ m = 2 - √3

So, the slope of line AB is 2 ± √3

So, the equations of other two lines joining the point (2, 3) are

y – 3 = 2 ± √3 (x – 2)

Hence, the given statement is TRUE

Given two lines are:

4x + y - 1 = 0 …(i)

and 7x - 3y -35 = 0 …(ii)

Now, point of intersection of these lines can be find out as:

Multiplying eq. (i) by 3, we get

12x + 3y – 3 = 0 …(iii)

On adding eq. (ii) and (iii), we get

7x – 3y – 35 + 12x + 3y – 3 = 0

⇒ 19x – 38 = 0

⇒ 19x = 38

⇒ x = 2

On putting value of x in (i), we get

4(2) + y – 1 = 0

⇒ 8 + y – 1 = 0

⇒ y = -7

So, the point of intersection of given two lines is:

(x, y) = (2, -7)

Now, we have to find the equation of the line joining the point (3, 5) and (2, -7).

Again, equation of line is given by:

⇒ y – 5 = 12(x – 3)

⇒ y – 5 = 12x – 36

⇒ 12x – y – 36 + 5 = 0

⇒ 12x – y – 31 = 0 …(iv)

Now, the distance of eq. (iv) from the point (0, 0) is

Now, the distance of eq. (iv) from the point (8, 34) is

Hence, the equation of line 12x – y – 31 = 0 is equidistant from (0, 0) and (8, 34)

Hence, the given statement is TRUE

We have equation of line

…(i)

Equation of line passing through the origin and perpendicular to the given line

…(ii)

Now, the foot of perpendicular from origin on the line (i) is the point of intersection of lines (i) and (ii).

So, to find its locus we have to eliminate the variable a and b.

Squaring and adding eq. (i) and (ii), we get

⇒ x² + y² = c²

Hence, the given statement is TRUE

Given that ax + 2y + 1 = 0,

bx + 3y + 1 = 0

and cx + 4y + 1 = 0 are concurrent

We know that,

It is given that the given lines are concurrent.

Now, expanding along first column, we get

⇒ a[3 – 4] – b[2 – 4] + c[2 – 3] = 0

⇒ – a + 2b – c = 0

⇒ 2b = a + c

and we know that, if a, b, c are in AP then

This means given lines are in AP not in GP.

Hence, the given statement is FALSE

Given points are (3, -4), (-2, 6), (-3, 6) and (9, -18)

Now, we find the slope because if the lines are perpendicular then the product of the slopes is -1 i.e. m₁m₂ = -1

Slope of the line joining the points (3, -4) and (-2, 6)

Here, x₁ = 3, x₂ = -2, y₁ = -4 and y₂ = 6

⇒ m₁ = -2

Now, slope of the line joining the points (-3, 6) and (9, -18)

Here, x₁ = -3, x₂ = 9, y₁ = 6 and y₂ = -18

⇒ m₂ = -2

∵ m₁ = m₂ = -2

and m₁m₂ = -2 × (-2) = -4 ≠ -1

So, the lines are parallel and not perpendicular

Hence, the given statement is FALSE

(a) Let P(x₁, y₁) be any point on the given line x + 5y = 13

∴ x₁ + 5y₁ = 13

⇒ 5y₁ = 13 – x₁ …(i)

Distance of the point P(x₁,y₁) from the equation 12x – 5y + 26 = 0

[given d = 2]

[from eq. (i)]

⇒ 2 = |x₁ + 1|

⇒ 2 = ±(x₁ + 1)

either x₁ + 1 = 2 or -(x₁ + 1) = 2

x₁ = 1 …(ii) or x₁ + 1 = -2

or x₁ = -3 …(iii)

Putting the value of x₁ = 1 in eq. (i), we get

5y₁ = 13 – 1

⇒ 5y₁ = 12

Putting the value of x₁ = -3 in eq. (i), we get

5y₁ = 13 – (-3)

⇒ 5y₁ = 13 + 3

⇒ 5y₁ = 16

Hence, the required points on the given line are

Hence, (a) ↔ (iii)

(b) Let P(x₁,y₁) be any point lying in the equation x + y = 4

∴ x₁ + y₁ = 4 …(i)

Distance of the point P(x₁,y₁) from the equation 4x + 3y = 10

[given]

⇒ 4x₁ + 3y₁ – 10 = ±5

either 4x₁ + 3y₁ – 10 = 5 or 4x₁ + 3y₁ – 10 = -5

4x₁ + 3y₁ = 5 + 10 or 4x₁ + 3y₁ = -5 + 10

4x₁ + 3y₁ = 15 …(ii) or 4x₁ + 3y₁ = 5 …(iii)

From eq. (i), we have y₁ = 4 – x₁ …(iv)

Putting the value of y₁ in eq. (ii), we get

4x₁ + 3(4 – x₁) = 15

⇒ 4x₁ + 12 – 3x₁ = 15

⇒ x₁ = 15 – 12

⇒ x₁ = 3

Putting the value of x₁ in eq. (iv), we get

y₁ = 4 – 3

⇒ y₁ = 1

Putting the value of y₁ = 4 – x₁ in eq. (iii), we get

4x₁ + 3(4 – x₁) = 5

⇒ 4x₁ + 12 – 3x₁ = 5

⇒ x₁ = 5 – 12

⇒ x₁ = - 7

Putting the value of x₁ in eq. (iv), we get

y₁ = 4 – (-7)

⇒ y₁ = 4 + 7

⇒ y₁ = 11

Hence, the required points on the given line are (3,1) and (-7,11)

Hence, (b) ↔ (i)

(c) Given that AP = PQ = QB

and given points are A (-2, 5) and B (3, 1)

Firstly, we find the slope of the line joining the points (-2, 5) and (3, 1)

Now, equation of line passing through the point (-2, 5)

y – y₁ = m(x – x₁)

⇒ 5y – 25 = -4(x + 2)

⇒ 5y – 25 = -4x – 8

⇒ 4x + 5y – 25 + 8 = 0

⇒ 4x + 5y – 17 = 0

Let P(x₁, y₁) and Q(x₂, y₂) be any two points on the AB

P(x₁, y₁) divides the line AB in the ratio 1:2

So, the coordinates of P(x₁, y₁) is

Now, Q(x₂, y₂) is the midpoint of PB

Hence, the coordinates of Q(x₂, y₂) is

Hence, (c) ↔ (ii)

(a) Given equation is (2x + 3y + 4) + λ (6x – y + 12) = 0

⇒ 2x + 3y + 4 + 6λx – λy + 12λ = 0

⇒ (2 + 6λ)x + (3 – λ)y + 4 + 12λ = 0 …(i)

If eq. (i) is parallel to y – axis, then

3 – λ = 0

⇒ λ = 3

Hence, (a) ↔ (iv)

(b) Given equation is (2x + 3y + 4) + λ (6x – y + 12) = 0

⇒ 2x + 3y + 4 + 6λx – λy + 12λ = 0 …(i)

⇒ (2 + 6λ)x + (3 – λ)y + 4 + 12λ = 0

⇒ (3 – λ)y = -4 – 12λ – (2 + 6λ)x

Since the above equation is in y = mx + b form.

So, the slope of eq. (i) is

Now, the second equation is 7x + y – 4 = 0 …(ii)

⇒ y = - 7x + 4

So, the slope of eq. (ii) is

m₂ = -7

Now, eq. (i) is perpendicular to eq. (ii)

∴ m₁m₂ = -1

⇒ (2 + 6λ) × 7 = - (3 – λ)

⇒ 14 + 42λ = λ – 3

⇒ 41λ = -3 – 14

⇒ 41λ = -17

Hence, (b) ↔ (iii)

(c) Given equation is (2x + 3y + 4) + λ (6x – y + 12) = 0

If the above equation passes through the point (1, 2) then

[2 × 1 + 3 × 2 + 4] + λ[6 × 1 – 2 + 12] = 0

⇒ 2 + 6 + 4 + λ (6 + 10) = 0

⇒ 12 + 16λ = 0

⇒ 12 = -16λ

Hence, (c) ↔ (i)

(d) Given equation is (2x + 3y + 4) + λ (6x – y + 12) = 0

⇒ 2x + 3y + 4 + 6λx – λy + 12λ = 0

⇒ (2 + 6λ)x + (3 – λ)y + 4 + 12λ = 0 …(i)

If eq. (i) is parallel to x – axis, then

2 + 6λ = 0

⇒ 6λ = -2

Hence, (d) ↔ (ii)

(a) Given equations are 2x – 3y = 0 …(i)

and 4x – 5y = 2 …(ii)

Equation of line passing through eq. (i) and (ii), we get

(2x – 3y) + λ(4x – 5y – 2 ) = 0 …(iii)

If the above eq. passes through the point (2, 1), we get

(2 × 2 – 3 × 1) + λ (4 × 2 – 5 × 1 – 2) = 0

⇒ (4 – 3) + λ (8 – 5 – 2) = 0

⇒ 1 + λ = 0

⇒ λ = -1

Putting the value of λ in eq. (iii), we get

(2x – 3y) + (-1)(4x – 5y – 2 ) = 0

⇒ 2x – 3y – 4x + 5y + 2 = 0

⇒ -2x + 2y + 2 = 0

⇒ x – y – 1 = 0

Hence, (a) ↔ (iii)

(b) Given equations are 2x – 3y = 0 …(i)

and 4x – 5y = 2 …(ii)

Equation of line passing through eq. (i) and (ii), we get

(2x – 3y) + λ(4x – 5y – 2 ) = 0 …(iii)

⇒ 2x – 3y + 4λx - 5λy - 2λ = 0

⇒ x(2 + 4λ) – y(3 + 5λ) – 2λ = 0

⇒ – y(3 + 5λ) = -(2 + 4λ) + 2λ

Since, the above equation is in y = mx + b form.

So, Slope of eq. (iii) is

Now, we find the slope of the given line x + 2y + 1 = 0 …(iv)

⇒ 2y = -x – 1

Since, the above equation is in y = mx + b form.

So, Slope of eq. (iv) is

We know that, if two lines are perpendicular to each other than the product of their slopes is equal to -1 i.e.

m₁m₂ = -1

Now, given that eq. (iii) is perpendicular to the given line x + 2y + 1 = 0

⇒ 2 + 4λ = 2 × (3 + 5λ)

⇒ 2 + 4λ = 6 + 10λ

⇒ 4λ – 10λ = 6 – 2

⇒ -6λ = 4

Putting the value of λ in eq. (iii), we get

⇒ 6x – 9y – 8x + 10y + 4 = 0

⇒ -2x + y + 4 =0

⇒ 2x – y – 4 = 0

⇒ 2x – y = 4

Hence, (b) ↔ (i)

(c) Given equations are 2x – 3y = 0 …(i)

and 4x – 5y = 2 …(ii)

Equation of line passing through eq. (i) and (ii), we get

(2x – 3y) + λ(4x – 5y – 2 ) = 0 …(iii)

⇒ 2x – 3y + 4λx - 5λy - 2λ = 0

⇒ x(2 + 4λ) – y(3 + 5λ) – 2λ = 0

⇒ – y(3 + 5λ) = -(2 + 4λ) + 2λ

Since, the above equation is in y = mx + b form.

So, Slope of eq. (iii) is

Now, we find the slope of the given line 3x – 4y + 5 = 0

⇒ 3x + 5 = 4y

Since, the above equation is in y = mx + b form.

So, the Slope of above equation is

Now, we know that if the two lines are parallel than there slopes are equal.

∴ m₁ = m₂

⇒ 4(2 + 4λ) = 3(3 + 5λ)

⇒ 8 + 16λ = 9 + 15λ

⇒ 16λ – 15λ = 9 – 8

⇒ λ = 1

Putting the value of λ in eq. (iii), we get

(2x – 3y) + (1)(4x – 5y – 2 ) = 0

⇒ 2x – 3y + 4x – 5y – 2 = 0

⇒ 6x – 8y – 2 = 0

⇒ 3x – 4y – 1 = 0

Hence (c) ↔ (iv)

(d) Given equations are 2x – 3y = 0 …(i)

and 4x – 5y = 2 …(ii)

Equation of line passing through eq. (i) and (ii), we get

(2x – 3y) + λ(4x – 5y – 2 ) = 0 …(iii)

⇒ 2x – 3y + 4λx - 5λy - 2λ = 0

⇒ x(2 + 4λ) – y(3 + 5λ) – 2λ = 0

⇒ – y(3 + 5λ) = -(2 + 4λ) + 2λ

Since, the above equation is in y = mx + b form.

So, Slope of eq. (iii) is

Given that the equation is equally inclined with axes, it means that the lines make the equal angles with both the coordinate axes. It will make an angle of 45° or 135° with x – axis.

So, the slope of two lines equally inclined to the axes are

m₂ = tan 45° and tan 135°

= 1 and tan (180 – 45°)

= 1 and -1

So,

⇒ 2 + 4λ = - (3 + 5λ) ⇒ 2 + 4λ = (3 + 5λ)

⇒ 2 + 4λ = - 3 – 5λ ⇒ 4λ – 5λ = 3 – 2

⇒ 4λ + 5λ = -3 – 2 ⇒ -λ = 1

⇒ 9λ = -5 ⇒ λ = -1

Putting the value of λ in eq. (iii), we get

⇒ 18x – 27y – 20x + 25y + 10 = 0

⇒ -2x – 2y + 10 = 0

⇒ x + y – 5 = 0

If λ = -1, then the required equation is

(2x – 3y) + (-1)(4x – 5y – 2) = 0

⇒ 2x – 3y – 4x + 5y + 2 = 0

⇒ -2x + 2y + 2 = 0

⇒ x – y – 1 = 0