Explore topic-wise InterviewSolutions in .

Three lines are said to be concurrent, if they pass through a common point.i.e., three lines are said to be concurrent if the point of intersection of two of them lies on the third line. 

Given:

Lines xy = 0 and x + y = 1 

To find: 

The coordinates of the orthocenter of the triangle formed by the lines xy = 0 and x + y = 1 

Explanation: 

The equation xy = 0 represents a pair of straight lines. 

The lines can be written separately in the following way: 

x = 0 … (1) 

y = 0 … (2) 

The third line is 

x + y = 1 … (3) 

Lines (1) and (2) are perpendicular to each other as they are coordinate axes. 

Therefore, the triangle formed by the lines (1), (2) and (3) is a right-angled triangle. 

Thus, the or the centre of the triangle formed by the given lines is the intersection of x = 0 and y = 0, which is (0, 0).

(A) is the correct answer. The equation of the line x – y + 3 = 0 can be rewritten as y = x + 3 ⇒ m = tan θ = 1 and hence θ = 45°.

The given equation of lines are 

3x + y – 2 = 0              (1) 

2x – y – 3 = 0              (2) 

px + 2y – 3 = 0            (3) 

Solving (1) and (2), we have  

(1) + (2) ⇒ 5x – 5 ⇒ x = 1 

(1) ⇒ 3 + y – 2 = 0 

⇒ y = – 1 

∴ (1, – 1) is the point of intersection of (1) and (2) putting (1, – 1) in equation (3), 

we get, 

p. 1 + 2. (– 1) – 3 = 0 

⇒ p – 5 = 0 

⇒ p = 5

The given lines are 

ax + by = c ⇒ y = \(-\frac{ax}{b}+\frac{c}{b},m_1=-\frac{a}{b}\) 

and a'x +b'y = c' ⇒ y = \(\frac{a'}{b'}x+\frac{c'}{b'},m_2=-\frac{a'}{b'}\) 

Since the line are perpendicular. 

∴ m₁m₂ = – 1 

⇒ (−\(\frac{a}{b}\) ) (−\(\frac{a'}{b'}\)) = − 1 

⇒ \(\frac{aa'}{bb'}=-1\) 

⇒ aa′ + bb′ = 0, which is the required condition.

Given: 

a, b, c are in A.P. 

To find: 

The coordinates of that point. 

Explanation: 

If a, b, c are in A.P., then

a + c = 2b 

⇒ a – 2b + c = 0

Comparing the coefficient of ax + by + c = 0 and a – 2b + c = 0, we get x = 1 and y = -2 

Hence, the coordinate of that point is (1, -2).

The equation of the line perpendicular to 3x + y = 3 is given below: 

x – 3y + λ = 0 

This line passes through (2, 2). 

2 – 6 + λ = 0 

⇒ λ = 4 

So, the equation of the line will be 

x – 3y + 4 = 0 

⇒ y = 13x + 43 

Hence, the y-intercept is \(\frac{4}{3}\).

Given, slope(m) = 2/3  and the line passes through (3, 5).

Equation of the line in slope point form is y-y₁ = m(x -x₁)

∴ The equation of the required line is y – 5 = 2/3 (x-3)

∴ 3 (y – 5) = 2 (x – 3) 

∴ 3y – 15 = 2x – 6 

∴ 2x – 3y + 9 = 0

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

Slope of perpendicular line = 1/3

So, the line passes through (2, 2) and having slope 1/3 is

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

⇒ 3y – 6 = x – 2

⇒ 3y = x + 4

y = 1/3x + 4/3 [∵y = mx + c]

So, the y – intercept is 4/3

Hence, the correct option is (d)

i. y = 2x – 4 

∴ 2x – y – 4 = 0 is the equation in ax + by + c 

= 0 form.

ii. y = 4 

∴ 0x + 1y – 4 = 0 is the equation in ax + by + c 

= 0 form.

iii. x/2 + y/4 = 1  

∴ 2x +y/4

∴ 2x + y – 4 = 0 is the equation in ax + by + c 

= 0 form.

iv. x/3 - y/2 = 0 

∴ 2x – 3y = 0 

∴ 2x – 3y + 0 = 0 is the equation in ax + by + c 

= 0 form.

Given: the equation is x/a + y/b = 1 

We know that,

General equation of line y = mx + c.

bx + ay = ab

ay = – bx + ab

y = -bx/a + b

The slope intercept form of the given line.

∴ Slope = – b/a and y – intercept = b

Given, slope(m) = 1/2 and the line passes through (3, – 2).

Equation of the line in slope point form is y-y₁ = m(x-x₁)

∴ The equation of the required line is [y-(-2)] = 1/2 (x-3)

∴ 2(y + 2)=x – 3 

∴ 2y + 4 = x – 3 

∴ x – 2y – 7 = 0

Given, A (2, 4) and B (1, 7) 

Slope of AB = \(\frac {7-4}{1-2} =\)-3 1-2

Since the required line is parallel to line AB, slope of required line (m) = slope of AB 

∴ m = – 3 and the required line passes through the origin. 

Equation of the line having slope m and passing through origin (0, 0) is y = mx. 

∴ The equation of the required line is y = – 3x

Given: the equation is  \(\frac{x}{a}+\frac{y}{b}=1\) 

Concept Used: General equation of line y = mx + c. 

Explanation: 

bx + ay = ab 

⇒ ay = - bx + ab 

⇒ y = \(-\frac{b}{a}\)x + b 

Hence, the slope intercept form of the given line. 

∴ Slope = - b/a and y - intercept = b

Given, A line which is parallel to x–axis and passing through (3, – 5) 

To Find: The equation of the line. 

Formula used: The equation of line is [y – y₁ = m(x – x₁)] 

Explanation: Here, The line is parallel to the x–axis, 

So, The parallel lines have equal slopes, 

And, the slope of x–axis is always 0, then 

The slope of line, m = 0 

Coordinates of line are (x₁, y₁) = (3, – 5) 

The equation of line = y – y₁ = m(x – x₁) 

By putting the values, we get 

y – (– 5) = 0(x – 3)

y + 5 = 0 

Hence, The equation of line is y + 5 = 0

Given p = 8, α = 225° 

Concept Used: 

Equation of line in normal form. 

Explanation: 

So, the equation of the line in normal form is 

Formula Used:

x cos α + y sin α = p x cos 225° + y sin 225° = 8 

We know, cos (180° + θ) = – cos θ , sin (180° + θ) = – sin θ 

⇒ – cos 45° – y sin 45° = 8

⇒ \(-\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=8\)

⇒ x + y + 8√2 = 0 

Hence, the equation of line in normal form is x + y + 8√2 = 0 

Given: 

x + y − 4 = 0, 2x − y + 3 = 0 and x − 3y + 2 = 0 

To find: Point of intersection of pair of lines. 

Concept Used: 

Point of intersection of two lines. 

Explanation: 

x + y − 4 = 0 … (1) 

2x − y + 3 = 0 … (2) 

x − 3y + 2 = 0 … (3) 

Solving (1) and (2) using cross - multiplication method:

\(\frac{x}{3-4},\frac{y}{-8-3}=\frac{1}{-1-2}\)

⇒ \(x=\frac{1}{3},y=\frac{11}{3}\)

 Solving (1) and (3) using cross - multiplication method:

\(\frac{x}{2-12},\frac{y}{-4-2}=\frac{1}{-3-1}\)

⇒ \(x=\frac{5}{2},y=\frac{3}{2}\)

Similarly, solving (2) and (3) using cross - multiplication method:

 \(\frac{x}{-2+9}=\frac{y}{3-4}=\frac{1}{-6+1}\) 

⇒ \(x=\frac{7}{5},y=\frac{1}{5}\)

Hence, the coordinates of the vertices of the triangle are 

 \(\Big(\frac{1}{3},\frac{11}{3}\Big)\) ,\(\Big(\frac{5}{2},\frac{3}{2}\Big)\) and \(\Big(-\frac{7}{5},\frac{1}{5}\Big)\)

Given: p = 4, α = 150° 

Concept Used: Equation of line in normal form. 

Explanation: 

So, the equation of the line in normal form is 

Formula Used: x cos α + y sin α = p 

x cos 150° + y sin 150° = 4 

cos (180° – θ) = – cos θ , sin (180° – θ) = sin θ 

⇒ x cos(180° – 30°) + y sin(180° – 30°) = 4 

⇒ – x cos 30° + y sin 30° = 4

⇒ \(\frac{\sqrt{3}x}{2}+\frac{y}{2}=4\)

⇒ √3x – y + 8 = 0 

Hence, the equation of line in normal form is √3 x – y + 8 = 0

B. y – x – 1 = 0

Explanation:

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 equation (i), we get

x – y + 1 = 0

Or y – x – 1 = 0

Hence, the correct option is (b)

Given: p = 5, α = 60° 

Concept Used: 

Equation of line in normal form. 

Explanation: 

So, the equation of the line in normal form is 

Formula Used: 

x cos α + y sin α = p 

x cos 60° + y sin 60° = 5

 ⇒ \(\frac{x}{2}+\frac{\sqrt{3y}}{2}=5\)

⇒ x + √3y = 10 

Hence, the equation of line in normal form is x + √3y = 10.

Given:

a = b and ab = 25

Let us find the equation of the line which cutoff intercepts on the axes.

∴ a² = 25

a = 5 [considering only positive value of intercepts]

By using the formula,

The equation of the line with intercepts a and b is x/a + y/b = 1 

x/5 + y/5 = 1

By taking LCM

x + y = 5
∴ The equation of line is x + y = 5

B. x + y = 5

Explanation:

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 equation (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:

p = 4, α = 30°

The equation of the line in normal form is given by

Using the formula,
x cos α + y sin α = p

Now, substitute the values, we get
x cos 30° + y sin 30° = 4

x√3/2 + y1/2 = 4
√3x + y = 8

∴ The equation of line in normal form is √3x + y = 8.

Inclination of the line, \(\theta\) = 150° 

∴ Slope, m = tan(150°) = tan(90° + 60°) 

= – Cot 60° 

= – \(\frac{1}{\sqrt{3}}\) 

Since the line passes through (3, – 5) 

∴ Equation of the line is (y – y₀) = m(x – x₀) 

⇒ y − (−5) = −\(\frac{1}{\sqrt{3}}\) (x − 3) 

⇒ √3(y + 5) = x – 3 

⇒ x + √3y + (5√3 − 3) = 0, 

is the required equation of the line.

(i) p = 5, α = 60°

Given:

p = 5, α = 60°

The equation of the line in normal form is given by

Using the formula,
x cos α + y sin α = p

Now, substitute the values, we get
x cos 60° + y sin 60° = 5

x/2 + √3y/2 = 5
x + √3y = 10

∴ The equation of line in normal form is x + √3y = 10.

(ii) p = 4, α = 150°

Given:

p = 4, α = 150°

The equation of the line in normal form is given by

Using the formula,
x cos α + y sin α = p

Now, substitute the values, we get

x cos 150° + y sin 150° = 4

cos (180° – θ) = – cos θ , sin (180° – θ) = sin θ

x cos(180° – 30°) + y sin(180° – 30°) = 4

– x cos 30° + y sin 30° = 4

–√3x/2 + y/2 = 4

-√3x + y = 8

∴ The equation of line in normal form is -√3x + y = 8.

Given: A line which is perpendicular and parallel to x–axis respectively and passing through (4, 3)

By using the formula,

The equation of line: [y – y₁ = m(x – x₁)]

Let us consider,

Case 1: When Line is parallel to x–axis

The parallel lines have equal slopes,

And, the slope of x–axis is always 0, then

The slope of line, m = 0

Coordinates of line are (x₁, y₁) = (4, 3)

The equation of line is y – y₁ = m(x – x₁)

Now substitute the values, we get

y – (3) = 0(x – 4)

y – 3 = 0

Case 2: When line is perpendicular to x–axis

The line is perpendicular to the x–axis, then x is 0 and y is – 1.

The slope of the line is, m = y/x

= -1/0

Coordinates of line are (x₁, y₁) = (4, 3)

The equation of line = y – y₁ = m(x – x₁)

Now substitute the values, we get

y – 3 = (-1/0) (x – 4)
x = 4

∴ The equation of line when it is parallel to x – axis is y = 3 and it is perpendicular is x = 4.

Given: Point P(x, y) is equidistant from points A(6, -1) and B(2, 3)

i.e., distance of P from A = distance of P from B

⇒ \(\sqrt{(x-6)^2+(y+1)^2}\) = \(\sqrt{(x-2)^2+(y-3)^2}\)

Squaring both sides,

⇒ (x – 6)² + (y – 1)² = (x – 2)² + (y – 3)²

⇒ x² – 12x + 36 + y² – 2y + 1 = x² – 4x + 4 + y² – 6y + 9

⇒ –12x + 36 + 2y + 1 = – 4x + 4 – 6y + 9

⇒ – 8x + 8y = –24

⇒ x – y = 3

Therefore, x – y = 3 is the required relation.

A line which is equidistant from the lines y = 10 and y = – 2 

To Find: The equation of the line 

Formula used: The equation of line is [y – y₁ = m(x – x₁)] 

Explanation: A line which is equidistant from, two other lines, 

So, the slopes must be the same . 

Therefore, The slope of line y = 10 and y = – 2 is 0, because lines are parallel to the x–axis. 

Since, The required line will pass from the midpoint of the line joining (0, – 2) and (0, 10) 

The Midpoint formula = \(\Big[\frac{x+x_1}{2},\frac{y+y_1}{2}\Big]\)

So, The coordinates of the point will be \(\Big[0,\frac{10-2}{2}\Big]\) (0, 4) 

Since The equation of the line is : 

y – 4 = 0(x – 0) 

y = 4 

Hence, The equation of the line is y = 4

Given: A line which makes an angle of 150° with the x–axis and cutting off an intercept at 2

By using the formula,

The equation of a line is y = mx + c

We know that angle, θ = 150°

The slope of the line, m = tan θ

Where, m = tan 150°

= -1/ √3

Coordinate of y–intercept is (0, 2)

The required equation of the line is y = mx + c

Now substitute the values, we get

y = -x/√3 + 2

√3y – 2√3 + x = 0

x + √3y = 2√3

∴ The equation of line is x + √3y = 2√3

(i) Here, The slope is 2 and the coordinates are (0, 3) 

Now, The required equation of line is 

y = mx + c 

y = 2x + 3 

(ii) Here, The slope is – 1/3 and the coordinates are (0, – 4) 

Now, The required equation of line is 

y = mx + c 

y = -\(\frac{1}{3}\) x – 4

3y + x = – 12 

(iii) Here, The slope is – 2 and the coordinates are (– 3, 0) Now, The required equation of line is y – y₁ = m (x – x₁)

y – 0 = – 2(x + 3)

y = – 2x – 6

2x + y + 6 = 0

The given lines are y = – 3 and y = 13.

So, the resulting line is also of the form y = a,

where

a = \(\frac{− 3 + 13}{2}\) = 5

∴ The resulting line is y = 5.

(i) With slope 2 and y – intercept 3

The slope is 2 and the coordinates are (0, 3)

Now, the required equation of line is

y = mx + c

Substitute the values, we get

y = 2x + 3

(ii) With slope – 1/ 3 and y – intercept – 4

The slope is – 1/3 and the coordinates are (0, – 4)

Now, the required equation of line is

y = mx + c

Substitute the values, we get

y = -1/3x – 4

3y + x = – 12

(iii) With slope – 2 and intersecting the x–axis at a distance of 3 units to the left of origin

The slope is – 2 and the coordinates are (– 3, 0)

Now, the required equation of line is y – y₁ = m (x – x₁)

Substitute the values, we get

y – 0 = – 2(x + 3)

y = – 2x – 6

2x + y + 6 = 0

Equation of a line parallel to y-axis is x = a

x = a

Since the line passes through the point (– 4, 3).

∴ – 4 = a ⇒ a = – 4.

∴ The equation of the line is x = – 4.

Equation of a line in intercepts form is

\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 …(i)

Given, a = 2 and b = 5

∴ (1) ⇒ \(\frac{x}{2}\) + \(\frac{y}{5}\) = 1

⇒ 5x + 2y – 10 = 0, is the required equation of the line.

Equation of a line parallel to X-axis is of the form y = k. 

Since the line passes through B(4, – 3), k = -3 

∴ The equation of the required line is y = – 3.

Equation of a line in two point form is

y − y₁ = \(\frac{y_2 − y_1}{x_2 − x_1}\) (x − x1) …(i)

Given, x₁ = 2, x₂ = 4, y₁ = 2, y₂ = – 6

⇒ y − 2 = \(\frac{-6 − 2}{4 − 2}\) (x − 2)

⇒ y – 2 = – 4 (x – 2)

⇒ 4x + y – 10 = 0, is the required equation of the line.

Equation of a line parallel to X-axis is of the form y = k.

Since the line passes through (2, 3), k = 3 

∴ The equation of the required line is y = 3.

For two lines to be perpendicular, their product of slope must be equal to -1.

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

slope = (y₂-y₁/x₂-x₁)

⇒ Slope of line AB is equal to

(5+5/-2-2) = (10/-4) = (-5/2) = -2.5

And the slope of line CD is equal to

(1-3/1-x) = (-2/1-x)

Their product must be equal to -1

the slope of line AB×Slope of line CD = -1

⇒ -2.5 x (-2/1-x) = -1

⇒ 5 = x-1

⇒x = 6

(i) If the slope of the line is zero it means 

M = tan θ 

M = tan 0

Since, m = 0 

So, The line is parallel to x - axis. 

(ii) If the slope of the line is positive it means Since θ is an acute 

So, The line makes an acute angle with the positive x - axis. 

(iii) If the slope of the line is positive it means 

Since, θ is an obtuse 

So, The line makes an obtuse angle with positive x - axis.

We have given coordinates of two lines (3, y) and (2, 7), (– 1, 4) and (0, 6) 

To Find: Value of y? 

The concept used: Slopes of the parallel line are always equal. 

The formula used: The slope of line = \(\frac{y_2-y_1}{x_2-x_1}\)

Now, The slope of the line whose coordinates are (3, y) and (2, 7). 

M₁ = \(\frac{7-y}{2-3}\)    …… (1) 

And, Now, The slope of the line whose coordinates are (– 1, 4) and (0, 6). 

M₂ = \(\frac{6-4}{0-(-1)}\)

M₂ =  \(\frac{2}{1}\)…… (2) 

On equating the equation (1) and (2), we get

\(\frac{7-y}{2-3}=\frac{2}{1}\)

7 – y = 2(– 1) 

– y = – 2 – 7 

Y = 9 

Hence, The value of y is 9.

10(x + 1)² + (x + 1)( y – 2) – 3(y – 2)² = 0 …(1) 

Put x + 1 = X and y – 2 = Y

∴ (1) becomes

10x² + xy – 3y² = 0

10x² + 6xy – 5xy – 3y² = 0

2x(5x + 3y) – y(5x + 3y) = 0

(2x – y)(5x + 3y) = 0

5x + 3y = 0 and 2x – y = 0

5x + 3y = 0

5(x + 1) + 3(y – 2) = 0

5x + 5 + 3y – 6 = 0

∴ 5x + 3y – 1 = 0

2x – y = 0

2(x + 1) – (y – 2) = 0

2x + 2 – y + 2 = 0

∴ 2x – y + 4 = 0

Equation of a line in slope-intercept form is y = mx + c

Also, c = 3 and m = 5

∴ (1) ⇒ y = 5x + 3, is the required equation of the line.

Slope of the line segment passing through (3, y) and (2, 7),

m₁ = \(\frac{7 − y}{2 − 3}\) = y – 7

And slope of the line passing through (– 1, 4) and (0, 6)

m₂ = \(\frac{6 − 4}{0 + 1}\) = 2

Since, the lines are parallel,

So m₁ = m₂

⇒ y – 7 = 2

⇒ y = 9

Given line is 6x + 3y – 5 = 0

⇒ 3y = – 6x + 5

⇒ y = – 2x + \(\frac{5}{3}\)

∴ Slope = – 2.

(x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)2 = 0

∴ (x – 2)² – 2(x – 2)(y + 1) – (x – 2)(y + 1) + 2(y + 1)² = 0

∴ (x – 2) [(x – 2) – 2(y + 1)] – (y + 1)[(x – 2) – 2(y + 1)] = 0

∴ (x – 2)(x – 2 – 2y – 2) – (y + 1)(x – 2 – 2y – 2) = 0

∴ (x – 2)(x – 2y – 4) – (y + 1)(x – 2y – 4) = 0

∴ (x – 2y – 4)(x – 2 – y – 1) = 0

∴ (x – 2y – 4)(x – y – 3) = 0

∴ the separate equations of the lines are

x – 2y – 4 = 0 and x – y – 3 = 0.

Alternative Method :

(x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)² = 0 … (1)

Put x – 2 = X and y + 1 = Y

∴ (1) becomes,

X² – 3XY + 2Y² = 0

∴ X² – 2XY – XY + 2Y² = 0

∴ X(X – 2Y) – Y(X – 2Y) = 0

∴ (X – 2Y)(X – Y) = 0

∴ the separate equations of the lines are

∴ X – 2Y = 0 and X – Y = 0

∴ (x – 2) – 2(y + 1) = 0 and (x – 2) – (y +1) = 0

∴ x – 2y – 4 = 0 and x – y – 3 = 0.

Given:

The equations of the lines are

2x − y + 3 = 0 … (1)

x + y + 2 = 0 … (2)

Let m₁ and m₂ be the slopes of these lines.

m₁ = 2, m₂ = -1

Let θ be the angle between the lines.
Then, by using the formula

tan θ = [(m₁ – m₂) / (1 + m₁m₂)]

= [(2 + 1) / (1 + 2)]

= 3
So,

θ = tan^-1 (3)

∴ The acute angle between the lines is tan^-1 (3).

(i) 3x + y + 12 = 0 and x + 2y – 1 = 0

Given:

The equations of the lines are

3x + y + 12 = 0 … (1)

x + 2y − 1 = 0 … (2)

Let m₁ and m₂ be the slopes of these lines.

m₁ = -3, m₂ = -1/2

Let θ be the angle between the lines.
Then, by using the formula

tan θ = [(m₁ – m₂) / (1 + m₁m₂)]

= [(-3 + 1/2) / (1 + 3/2)]

= 1

So,

θ = π/4 or 45^o

∴ The acute angle between the lines is 45°

(ii) 3x – y + 5 = 0 and x – 3y + 1 = 0

Given:

The equations of the lines are

3x − y + 5 = 0 … (1)

x − 3y + 1 = 0 … (2)

Let m₁ and m₂ be the slopes of these lines.

m₁ = 3, m₂ = 1/3

Let θ be the angle between the lines.
Then, by using the formula

tan θ = [(m₁ – m₂) / (1 + m₁m₂)]

= [(3 + 1/3) / (1 + 1)]

= 4/3

So,

θ = tan^-1 (4/3)

∴ The acute angle between the lines is tan^-1 (4/3).

Given: 

The equation is perpendicular to √3x – y + 5 = 0 equation and cuts off an intercept of 4 units with the negative direction of y-axis.

The line perpendicular to √3x – y + 5 = 0 is x + √3y + λ = 0  

It is given that the line x + √3y + λ = 0 cuts off an intercept of 4 units with the negative direction of the y-axis.

This means that the line passes through (0,-4).

So,

Let us substitute the values in the equation x + √3y + λ = 0, we get

0 – √3 (4) + λ = 0 
λ = 4√3

Now, substitute the value of λ back, we get

x + √3y + 4√3 = 0

∴ The required equation of line is x + √3y + 4√3 = 0.

It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as y = m(x – d)
For the line intersecting the x-axis at a distance of 3 units to the left of the origin,
d = –3.
The slope of the line is given as m = –2
Thus, the required equation of the given line is y = –2 [x – (–3)] y = –2x – 6
i.e., 2x + y + 6 = 0

Given: A (a, b) and B (a₁, b₁) be the given points 

To find: 

Equation of the right bisector of the line segment joining the points (a, b) and (a₁, b₁). 

Explanation: 

Let C be the midpoint of AB.

∴ coordinates of C = \(\Big(\frac{a+a_2}{2},\frac{b+b_1}{2}\Big)\) 

And, slope of AB = \(\frac{b_1-b}{a_1-a}\) 

So, the slope of the right bisector of AB is \(-\frac{a_1-a}{b_1-b}\) 

Thus, the equation of the right bisector of the line segment joining the points (a, b) and (a₁, b₁) is

\(y - \frac{b+b_1}{2}\) = \(-\frac{a_1-a}{b_1-b}\Big(x-\frac{a+a_1}{2}\Big)\) 

⇒ 2 (a₁ - a)x + 2y(b₁ - b) + (a² + b²) – (a₁² + b₁²) = 0 

Hence, equation of the required line 2 (a₁ – a)x + 2y(b₁- b) + (a² + b²) – (a₁² + b₁²) = 0