The Straight Lines

Given

To Find: Slope of the line

Angle made with the positive x - axis is

The Slope of the line is m

Formula Used: m = tanθ

So, The slope of Line is m = tan = – 1

Hence, The slope of the line is – 1.

Given

To Find: Slope of the line

Angle made with the positive x - axis is

The Slope of the line is m

Formula Used: m = tanθ

So, The slope of Line is m = tan

Hence, The slope of the line is .

Given

To Find: Slope of the line

Angle made with positive x - axis is

The Slope of the line is m

Formula Used: m = tanθ

So, The slope of Line is m = tan

Hence, The slope of the line is – 1.

Given

To Find: Slope of the line

Angle made with positive x - axis is

The Slope of the line is m

Formula Used: m = tan θ

So, The slope of Line is m = tan

Hence, The slope of the line is √3.

Given (– 3, 2) and (1, 4)

To Find The slope of the line passing through the given points.

Here,

The formula used: Slope of line =

So, The slope of the line, m =

m =

Hence, The slope of the line is

Given (at²₁, 2at₁) and (at²₂, 2at₂)

To Find: The slope of the line passing through the given points.

The formula used: Slope of line =

So, The slope of the line, m =

m =

m =

[Since, (a^{2 –} b² = (a – b)(a + b)]

m =

Hence, The slope of the line is

Given (3, – 5) and (1, 2)

To Find: The slope of line passing through the given points.

Here,

The formula used: Slope of line =

So, The slope of the line, m =

m =

Hence, The slope of the line is

We have given Coordinates off two lines.

Given: (5, 6) and (2, 3); (9, – 2) and 96, – 5)

To Find: Check whether Given lines are perpendicular to each other or parallel to each other.

Concept Used: If the slopes of this line are equal the lines are parallel to each other. Similarly, If the product of the slopes of this two line is – 1, then lines are perpendicular to each other.

The formula used: Slope of a line, m =

Now, The slope of the line whose Coordinates are (5, 6) and (2, 3)

So, m₁ = 1

Now, The slope of the line whose Coordinates are (9, – 2) and (6, – 5)

So, m₂ = 1

Here, m_{1 =} m_{2 =} 1

Hence, The lines are parallel to each other.

We have given Coordinates off two line.

Given: (9, 5) and (– 1, 1); through (3, – 5) and (8, – 3)

To Find: Check whether Given lines are perpendicular to each other or parallel to each other.

Concept Used: If the slopes of this line are equal the the lines are parallel to each other. Similarly, If the product of the slopes of this two line is – 1, then lines are perpendicular to each other.

The formula used: Slope of a line, m =

Now, The slope of the line whose Coordinates are (9, 5) and (– 1, 1)

So, m₁ =

Now, The slope of the line whose Coordinates are (3, – 5) and (8, – 3)

So, m₂ =

Here, m₁ = m₂ =

Hence, The lines are parallel to each other.

We have given Coordinates off two line.

Given: (6, 3) and (1,1) and (– 2, 5) and (2, – 5)

To Find: Check whether Given lines are perpendicular to each other or parallel to each other.

Concept Used: If the slopes of this line are equal the the lines are parallel to each other. Similarly, If the product of the slopes of this two line is – 1, then lines are perpendicular to each other.

The formula used: Slope of a line, m =

Now, The slope of the line whose Coordinates are (6, 3) and (1, 1)

So, m₁ =

Now, The slope of the line whose Coordinates are (– 2, 5) and (2, – 5)

So, m₂ =

Here, m₁m_{2 =}

m₁m_{2 =} – 1

Hence, The line is perpendicular to other.

We have given Coordinates off two line.

Given: (3, 15) and (16, 6) and (– 5, 3) and (8, 2)

To Find: Check whether Given lines are perpendicular to each other or parallel to each other.

Now,

Concept Used: If the slopes of these line are equal the the lines are parallel to each other. Similarly, If the product of the slopes of these two line is – 1, then lines are perpendicular to each other.

The formula used: Slope of a line, m =

Now, The slope of the line whose Coordinates are (3, 15) and (16, 6)

So, m₁ =

Now, The slope of the line whose Coordinates are (– 5, 3) and (8, 2)

So, m₂ =

Here, m₁≠m₂ nor m₁m₂ = – 1

Hence, The lines are neither perpendicular and nor parallel to each other.

(i) Given, Line bisects the first quadrant

To Find: Find the slope of the line.

Here, If the line bisects in the first quadrant, then the angle must be between line and the positive direction of x - axis .

Since, Angle =

The formula used: The slope of the line, m = tan θ

Similarly, The slope of the line for a given angle is m = tan 45

m = 1

Hence, The slope of the line is 1.

(ii) To Find: Find the slope of the line.

Here, The line makes an angle of 30^o with the positive direction of y - axis (Given)

Since Angle between line and positive side of axis = 90^o + 30^o = 120^o

The formula used: The slope of the line, m = tan θ

Similarly, The slope of the line for a given angle is m = tan 120^o

m = – √3

Hence, The slope of the line is – √3 .

We have three points given A(4, 8), b(5, 12), C(9, 28)

To Prove: Given Points are collinear

Proof: A(4, 8), B(5, 12), C(9, 28)

The formula used: The slope of the line =

The slope of line AB =

AB =

The slope of line BC =

BC = = 4

The slope of line CA =

Here, AB = BC = CA

Hence, The Given points are collinear.

We have three points given A(16, – 18), B(3, – 6), C(– 10, 6)

To Prove: Given Points are collinear

Proof: AB[(16, – 18),(3, – 6)], BC[(3, – 6),(– 10, 6)], CA[(– 10, 6),(16, – 18)]

Formula used: The slope of the line =

The slope of line AB =

AB =

The slope of line BC =

BC =

The slope of line CA =

CA =

Here, AB = BC = CA

Hence, The Given points are collinear.

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 =

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

M₁ = …… (1)

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

M₂ =

M_{2 =} …… (2)

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

7 – y = 2(– 1)

– y = – 2 – 7

Y = 9

Hence, The value of y is 9.

(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 .

To Prove: The given line is parallel to another line.

Proof: Let Assume the coordinate A(2, – 3) and B(– 5, 1), C(7, – 1) and D(0,3).

The concept used: Slopes of the parallel lines are equal.

The formula used: The slope of the line, m =

Now, The slope of AB =

The Slope of AB =

Now, The slope of CD =

The Slope of AB =

So, The slope of AB = The slope of CD

Hence, The given Lines are parallel to each other.

To Prove: The Given line is perpendicular to each other.

Proof: Let Assume the coordinate A(2, – 5) and B(– 2, 5) joining the line AB, C(6,3) and D(1,1) joining the line CD.

The concept used: The product of the slopes of lines always – 1.

The formula used: The slope of the line, m =

Now, The slope of AB =

The Slope of AB =

Now, The slope of CD =

The Slope of AB =

So, AB × CD =

AB × CD = – 1

Hence, The given Lines are perpendicular to each other.

We have given three points of a triangle.

Given: A(0, 4), B(1, 2), C(3, 3)

To Prove: Given points are the vertices of Right – angled Triangle.

Proof: We have A(0, 4), B(1, 2), C(3, 3)

The concept used: If the two lines are perpendicular to each other then it will be a right – angled triangle.

Now, Joining the points to make a line as AB, BC, and CA

The formula used: The slope of the line, m =

Now, The slope of line m_AB =

The slope of m_AB =

and, The slope of line BC =

The slope of m_Bc =

Now, m_AB × m_Bc =

m_AB × m_Bc = – 1

Since, AB is perpendicular to BC, it means B =

Hence, ABC is a right angle Triangle.

To Prove: Given vertices are of the rectangle.

Explanation: We have given points A(– 4, – 1), B(– 2, – 4), C(4, 0) and D(2, 3)

The points are joining in the form of AB, BC, CD, and AD

The formula used: The slope of the line, m =

Now, The slope of Line AB, m_AB

m_AB =

The slope of BC, m_BC =

m_{BC =}

Now, The slope of Line CD, m_CD =

m_CD =

The slope of AD, m_AD =

m_{AD =}

Here, We can see that, m_{AB =} m_CD and m_{BC =} m_AD

i.e, AB CD and BC AD

And, m_AB×m_BC =

M_CD×m_AD =

So, that ABBC and CDAD

Hence, ABCD is a Rectangle.

If these three points lie on a line, the slope will be equal.

So, slope of A(h, 0) and P(a, b) = Slope of A(h, 0) and B(0, k)

Slope of AP

Slope of AB

Now,

bh = - ka + kh

ak + bh = kh

Dividing both sides by kh, we get,

Given, The tangent of the angle between them is

To Find Slope of the other line.

Assumption: The slope of line m₁ = x, and m₂ = 2x

Formula used:

Explanation: We have tan given, then

Case 1:

2x² + 1 = 3x – 6x

2x² + 3x + 1 = 0

2x² + 2x + x + 1 = 0

2x(x + 1) + 1(x + 1) = 0

(2x + 1)(x + 1) = 0

x = – 1,

Case 2:

2x² + 1 = 3x

2x² – 3x + 1 = 0

2x^{2 –} 2x – x + 1 = 0

2x(x – 1) – 1(x – 1) = 0

(2x – 1)(x – 1) = 0

x = 1,

Hence, The slope of other line is either 1, or – 1, .

For the given graph,

Slope of line AB

Now, Slope of AB = Slope of AC

Therefore,

Slope of AC

5p – 460 = 25

5p = 485

P = 97 Crores.

To Prove: Given points are of Parallelogram.

Explanation: Let us Assume that we have points, A (– 2, – 1), B(4, 0), C(3, 3) and D(– 3, 2), are joining the sides as AB, BC, CD, and AD.

The formula used: The slope of the line, m =

Now, The slope of Line AB, m_AB =

m_AB =

The slope of BC, m_BC =

m_{BC =}

Now, The slope of Line CD, m_CD =

m_CD =

The slope of AD, m_AD =

m_{AD =}

Here, We can see that, m_{AB =} m_CDand m_BC= m_AD

i.e, AB CD and BC AD

We know, If opposite side of a quadrilateral are parallel that it is parallelogram.

Hence, ABCD is a Parallelogram.

Given, (3, – 1) and (4, – 2)

To find: Find the angle between x - axis and the line.

Explanation: We have two points A(3, – 1) and B(4, – 2).

The formula used: The slope of the line, m =

Now, The slope of line AB, m_AB =

m_{AB =} – 1

and, we know, The slope of x - axis is always 0

Now, the angle between x - axis and slope of line AB is,

θ = tan ^{– 1} – 1

θ = 135^o

Hence, The angle between the x - axis and the line is 135⁰.

To Find: Find the value of x ?

The concept used: If two line is perpendicular then, the product of their slopes is – 1.

Explanation: We have two lines having point A(– 2,6) and B(4,8) and other line having points C(8,12) and D(x,24).

The formula used: The slope of the line, m =

Now, The slope of Line AB is, m_AB =

m_AB =

and, The slope of Line CD is, m_CD =

m_CD =

We know the product of the slopes of perpendicular line is always – 1. Then,

m_AB × m_CD = – 1

= – 1

x – 8 = – 4

x = – 4 + 8

x = 4

Hence, The value of x is 4.

The given points (x, – 1), (2, 1) and (4, 5) are collinear.

To Find: The value of x.

Concept Used: It is given that points are collinear, SO the area of the triangle formed by the points must be zero.

Formula used: The area of triangle = x₁(y₁ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)

Explanation: Let be points of triangle A(x, – 1), B(2, 1) and C(4, 5)

Now, The points are collinear than, Area of a triangle is zero.

Here, Put the given values in formula and we get,

x(1 – 5) + (2)(5 – (– 1)) + 4(– 1 – 1) = 0

x – 5x + 12 – 8 = 0

– 4x + 4 = 0

4x = 4

x = 1

Hence, The value of x is 1.

Given, (3, – 1) and (4, – 2)

To find: Find the angle between x - axis and the line.

Explanation: We have two points A(3, – 1) and B(4, – 2).

The formula used: The slope of the line, m =

Now, The slope of line AB, m_AB =

m_{AB =} – 1

and, we know, The slope of x - axis is always 0

Now, the angle between x - axis and slope of line AB is,

θ = tan ^{– 1} – 1

θ = 135^o

Hence, The angle between the x – axis and the line is 135⁰.

To Prove: Given points are of Parallelogram.

Explanation: Let us Assume that we have points, A (– 2, – 1), B(4, 0), C(3, 3) and D(– 3, 2), are joining the sides as AB, BC, CD, and AD.

The formula used: The slope of the line, m =

Now, The slope of Line AB, m_AB =

m_AB =

The slope of BC, m_BC =

m_{BC =}

Now, The slope of Line CD, m_CD =

m_CD =

The slope of AD, m_AD =

m_{AD =}

Here, We can see that, m_{AB =} m_CDand m_BC= m_AD

i.e, AB CD and BC AD

We know, If opposite side of a quadrilateral are parallel that it is a parallelogram.

Hence, ABCD is a Parallelogram.

Given, A quadrilateral has vertices (4, 1), (1, 7), (– 6, 0) and (– 1, – 9).

To Prove: Mid – Points of the quadrilateral form a parallelogram.

The formula used: Mid point formula =

Explanation: Let ABCD is a quadrilateral

E is the midpoint of AB

F is the midpoint of BC

G is the midpoint of CD

H is the midpoint of AD

Now, Find the Coordinates of E, F,G and H using midpoint Formula

Coordinate of E =

Coordinate of F =

Coordinate of G =

Coordinate of H =

Now, EFGH is a parallelogram if the diagonals EG and FH have the same mid – point

Coordinate of mid – point of EG =

Coordinate of mid – point of FH =

Since Diagonals are equals then EFGH is a parallelogram.

Hence, EFGH is a parallelogram.

Given:

The equations of the lines are as follows:

2x − y + 3 = 0 … (1)

x + y − 5 = 0 … (2)

Concept Used:

Point of intersection of two lines.

To find:

Point of intersection of pair of lines.

Explanation:

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

⇒ x = 2/3 and y = 13/3

Hence, the point of intersection is

Given:

The equations of the lines are as follows:

bx + ay = ab

To find:

Point of intersection of pair of lines.

Concept Used:

Point of intersection of two lines.

Explanation:

⇒ bx + ay − ab = 0 … (1)

ax + by = ab ⇒ ax + by − ab = 0 … (2)

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

and

Hence, the point of intersection is

Given:

The equations of the lines are

and

To find:

Point of intersection of pair of lines.

Concept Used:

Point of intersection of two lines.

Explanation:

Thus, we have:

………(1)

……….(2)

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

,

and

Hence, the point of intersection is or

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:

⇒ x = 1/3, y = 11/3

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

⇒ x = 5/2, y = 3/2

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

⇒ x = - 7/5, y = 1/5

Hence, the coordinates of the vertices of the triangle are

Given:

y (t₁ + t₂) = 2x + 2a t₁t₂, y (t₂ + t₃) = 2x + 2a t₂t₃ and y (t₃ + t₁) = 2x + 2a t₁t₃

To find:

Point of intersection of pair of lines.

Concept Used:

Point of intersection of two lines.

Explanation:

2x − y (t₁ + t₂) + 2a t₁t₂ = 0 … (1)

2x − y (t₂ + t₃) + 2a t₂t₃ = 0 … (2)

2x − y (t₃ + t₁) + 2a t₁t₃ = 0 … (3)

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

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

Solving (2) and (3) using cross - multiplication method:

Hence, the coordinates of the vertices of the triangle are

Given:

y = m₁x + c₁ … (1)

y = m₂x + c₂ … (2)

x = 0 … (3)

Explanation:

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively. Solving (1) and (2):

Thus, AB and BC intersect at B

Solving (1) and (3):

x = 0, y = c₁

Thus, AB and CA intersect at A 0,c₁.

Similarly, solving (2) and (3):

x = 0, y = c₂

Thus, BC and CA intersect at C 0,c₂.

∴ Area of triangle ABC =

Given:

y = 0 … (1)

x = 2 … (2)

x + 2y = 3 … (3)

Assuming:

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Concept Used:

Point of intersection of two lines.

Explanation:

Solving (1) and (2):

x = 2, y = 0

Thus, AB and BC intersect at B (2, 0).

Solving (1) and (3):

x = 3, y = 0

Thus, AB and CA intersect at A (3, 0).

Similarly, solving (2) and (3):

x = 2, y = 12

Thus, BC and CA intersect at C(2, 12).

∴ Area of triangle ABC =

Hence, area of triangle ABC is .

Given:

x + y − 6 = 0 … (1)

x − 3y − 2 = 0 … (2)

5x − 3y + 2 = 0 … (3)

Assuming:

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Concept Used:

Point of intersection of two lines.

Explanation:

Solving (1) and (2):

x = 5, y = 1

Thus, AB and BC intersect at B (5, 1).

Solving (1) and (3):

x = 2, y = 4

Thus, AB and CA intersect at A (2, 4).

Similarly, solving (2) and (3):

x = −1, y = −1

Thus, BC and CA intersect at C (−1, −1).

∴ Area of triangle ABC =

Hence, area of triangle ABC is .

Given: equations are as follows:

3x + 2y + 6 = 0 … (1)

2x − 5y + 4 = 0 … (2)

x − 3y − 6 = 0 … (3)

Assuming:

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Concept Used:

Point of intersection of two lines.

Explanation:

Solving (1) and (2):

x = −2, y = 0

Thus, AB and BC intersect at B (−2, 0).

Solving (1) and (3):

x = - 6/11, y = - 24/11

Thus, AB and CA intersect at

Similarly, solving (2) and (3):

x = −42, y = −16

Thus, BC and CA intersect at C (−42, −16).

Let D, E and F be the midpoints the sides BC, CA and AB, respectively. Then,

Then, we have:

Now, the equation of the median AD is

⇒ 16x – 59y – 120 = 0

The equation of the median BE is

⇒ 25x – 53y + 50 = 0

And, the equation of median CF is

⇒ 41 x – 112 y – 70 = 0

Given: equations are as follows:

y = √3 x + 1……(1)

y = 4 …….(2)

y = - √3 x + 2…….(3)

Assuming:

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

To prove:

Lines y = √3 x + 1, y = 4 and y = - √3 x + 2 form an equilateral triangle.

Explanation:

Solving (1) and (2):

x = √3 , y = 4

Thus, AB and BC intersect at B(√3,4)

Solving (1) and (3):

Thus, AB and CA intersect at A(,)

Similarly, solving (2) and (3):

, y = 4

Thus, BC and AC intersect at C(,4)

Now, we have:

Hence Proved, the given lines form an equilateral triangle

Let a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 be the two lines.

(a) The lines intersect if is true.

(b) The lines are parallel if is true.

(c) The lines are coincident if is true.

(i) Given: 2x + y − 1 = 0 and 3x + 2y + 5 = 0

Explanation:

Here,

Therefore, the lines 2x + y − 1 = 0 and 3x + 2y + 5 = 0 intersect.

(ii) Given: x − y = 0 and 3x − 3y + 5 = 0

Explanation:

Here,

Therefore, the lines x − y = 0 and 3x − 3y + 5 = 0 are parallel.

(iii) Given: 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0

Explanation:

Here,

Therefore, the lines 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0 are coincident.

Given:

4x + y − 1 = 0 … (1)

7x − 3y − 35 = 0 … (2)

Concept Used:

Point of intersection of two lines.

Explanation:

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

⇒ x = 2 ,y = - 7

Thus, the point of intersection of the given lines is (2, - 7).

So, the equation of the line joining the points (3, 5) and (2, - 7) is

⇒ y - 5 = 12x - 36

⇒ 12x – y – 31 = 0

Hence, the required equation of line is 12x – y – 31 = 0

Given:

4x − 7y − 3 = 0 … (1)

2x − 3y + 1 = 0 … (2)

To find:

Equation of line passing through the point of intersection of lines.

Concept Used:

Point of intersection of two lines.

Explanation:

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

⇒ x = - 8 , y = - 5

Thus, the point of intersection of the given lines is ( - 8, - 5).

Now, the equation of a line having equal intercept as a is

This line passes through ( - 8, - 5)

⇒ - 8 - 5 = a

⇒ a = - 13

Hence, the equation of the required line is or x + y + 13 = 0

Given: lines are as follows:

y = m1 x … (1)

y = m2 x … (2)

y = c … (3)

To prove:

The area of the triangle formed by the lines y = m₁x, y = m₂x and y = c is equal to .

Concept Used:

Point of intersection of two lines.

Explanation:

Solving (1) and (2), we get (0, 0) as their point of intersection.

Solving (1) and (3), we get as their point of intersection.

Similarly, solving (2) and (3), we get as their point of intersection.

∴ Area of the triangle formed by these lines =

It is given that m1 and m2 are the roots of the

∴ AREA =

Hence Proved.

Given: lines are x + y = 3 and 2x − 3y = 1.

To find:

a and b.

Concept Used:

Point of intersection of two lines.

Explanation:

x + y − 3 = 0 … (1)

2x − 3y − 1 = 0 … (2)

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

⇒ x = 2 , y = 1

Thus, the point of intersection of the given lines is (2, 1).

It is given that the line passes through (2, 1).

∴ ……(3)

It is also given that the line is parallel to the line x − y − 6 = 0.

Hence, Slope of is equal to the slope of x − y − 6 = 0 or, y = x – 6

∴ - b/a = 1

⇒ b = - a … (4)

From (3) and (4):

∴

⇒ a = 1

From (4):

b = −1

∴ a = 1,

b = −1

Hence, a = 1, b = - 1

Given: Sides of triangle are are x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0.

Assuming: AB, BC and AC be the sides of triangle whose equation is are x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0.

To find:

Orthocenter of triangle.

Concept Used:

Point of intersection of two lines.

Explanation:

x + y – 1 = 0 …… (i)

2x + 3y – 6 = 0 …… (ii)

4x – y + 4 = 0. …… (iii)

By solving equation (i) and (ii) By cross multiplication

⇒ x = - 3 ,y = 4

B( - 3, 4)

By Solving equation (i) and (iii) By cross multiplication

⇒ x , y

A

Equation of BC is 2x + 3y = 6

Altitude AD is perpendicular to BC,

Therefore, equation of AD is x + y + k = 0

AD is passing through A

⇒ k = - 1

Equation of AD is x + y – 1 = 0 …… (iv)

Altitude BE is perpendicular to AC.

⇒ Let the equation of DE be x – 2y = k

BE is passing through D( - 3, 4)

⇒ - 3 – 8 = k

⇒ k = - 11

Equation of BE is x – 2y = - 11 …… (v)

By solving equation (iv) and (v),

We get, x = - 3 and y = 4

Hence, the orthocenter of triangle is ( - 3, 4).

Given:

The sides AB, BC and CA of a triangle ABC are as follows:

5x − 3y + 2 = 0 … (1)

x − 3y − 2 = 0 … (2)

x + y − 6 = 0 … (3)

To find:

The equation of the Altitude through the vertex A.

Concept Used:

Point of intersection of two lines.

Explanation:

Solving (1) and (3):

x = 2 , y = 4

Thus, AB and CA intersect at A (2, 4).

Let AD be the altitude.

AD⊥BC

∴ Slope of AD × Slope of BC = −1

Here, slope of BC = slope of the line (2) = 1/3

∴ Slope of AD×1/3 = - 1 ⇒ Slope of AD = - 3

Hence, the equation of the altitude AD passing through A (2, 4) and having slope −3 is y - 4 = - 3x - 2⇒ 3x + y = 10

Given: coordinates of the orthocenter of the triangle whose vertices are ( - 1, 3), (2, - 1) and (0, 0).

Assuming:

A (0, 0), B (−1, 3) and C (2, −1) be the vertices of the triangle ABC.

Let AD and BE be the altitudes.

To find:

Orthocenter of the triangle.

Explanation:

AD⊥BC and BE⊥AC

∴ The slope of AD × Slope of BC = −1

The slope of BE × Slope of AC = −1

Here, the slope of BC =

and slope of AC =

∴ slope of AD × ( - 4/3) = - 1 and slope of BE × ( - 1/2) = - 1

⇒ slope of AD and slope of BE = 2

The equation of the altitude AD passing through A (0, 0) and having slope is

y - 0 ( x - 0)

⇒ y x …..(1)

The equation of the altitude BE passing through B (−1, 3) and having slope 2 is

y - 3 = 2(x + 1)

⇒ 2x – y + 5 = 0 …….(2)

Solving (1) and (2):

x = − 4, y = − 3

Hence, the coordinates of the orthocentre is (−4, −3).

Given: lines are as follows:

3x − 4y = 0 … (1)

12y + 5x = 0 … (2)

y − 15 = 0 … (3)

Assuming:

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Concept Used:

Point of intersection of two lines.

Explanation:

Solving (1) and (2):

x = 0, y = 0

Thus, AB and BC intersect at B (0, 0).

Solving (1) and (3):

x = 20 , y = 15

Thus, AB and CA intersect at A (20, 15).

Solving (2) and (3): x = −36 , y = 15

Thus, BC and CA intersect at C (−36, 15).

Let us find the lengths of sides AB, BC and CA.

Here, a = BC = 39, b = CA = 56 and c = AB = 25

Also, x₁, y₁ = A (20, 15), x₂, y_{2 =} B (0, 0) and x₃, y₃ = C (−36, 15)

AND incentre

Hence, coordinate of incenter and centroid are ( - 1, 8)

Given: lines are as follows:

To prove:

lines form a rhombus.

Assuming:

In quadrilateral ABCD, let equations (1), (2), (3) and (4) represent the sides AB, BC, CD and

DA, respectively.

Explanation:

Lines (1) and (3) are parallel and lines (2) and (4) are parallel.

Solving (1) and (2):

x = 0, y = 0.

Thus, AB and BC intersect at B (0, 0).

Solving (1) and (4):

x , y

Thus, AB and DA intersect A

Solving (3) and (2):

x , y =

Thus, BC and CD intersect at C

Solving (3) and (4):

x , y

Thus, DA and CD intersect at D

Let us find the lengths of sides AB, BC and CD and DA.

AB

BC

CB

DA = 1

Hence Proved, the given lines form a rhombus.

Given: Line passing through point of intersection of lines 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.

To find:

The equation of the required line.

Concept Used:

Point of intersection of two lines.

Explanation:

2x + y – 5 = 0 …… (i)

x + 3y + 8 = 0 …… (ii)

By solving equation (i) and (ii) ,By cross multiplication,

⇒ x , y

Point of intersection

Equation of line parallel to 3x + 4y – 7 = 0 is

3x + 4y + k = 0 …… (iii)

Equation (iii) passing through point of intersection of line.

⇒

⇒ k = 3

Hence, required equation of line is 3x + 4y + 3 = 0.

Given: the point of intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x – 5y + 11 = 0.

To find:

The equation of the required line.

Concept Used:

Point of intersection of two lines.

Explanation:

5x – 6y – 1 = 0 …… (i)

3x + 2y + 5 = 0 …… (ii)

By solving equation (i) and (ii) ,By cross multiplication,

⇒ x , y

Point of intersection ( - 1, - 1)

Now, the slope of the line 3x – 5y + 11 = 0 or yis

Now, we know that rhe product of the slope of two perpendicular lines is - 1.

Assuming: the slope of required line is m

⇒

Now, the equation of the required line passing through ( - 1, - 1) and having slope is given by,

Y + 1

⇒ 3y + 3 = - 5x – 5

⇒ 5x + 3y + 8 = 0

Hence, equation of required line is 5x + 3y + 8 = 0.

Given:

15x – 18y + 1 = 0 …… (i)

12x + 10y – 3 = 0 …… (ii)

6x + 66y – 11 = 0 …… (iii)

To prove:

Sets of given three lines are concurrent.

Explanation:

Now, consider the following determinant:

= 15(- 110 + 198) + 18(-132 + 18) + 1(792 – 60)

⇒ 1320 – 2052 + 732 = 0

Hence proved, the given lines are concurrent.

Given:
3x − 5y − 11 = 0 …… (i)

5x + 3y − 7 = 0 …… (ii)

x + 2y = 0 …… (iii)

To prove:

Sets of given three lines are concurrent.

Explanation:

Now, consider the following determinant:

= 3 × 14 + 5 × 7 – 11 × 7 = 0

Hence, the given lines are concurrent.

Given:
bx + ay – ab = 0 … (1)

ax + by – ab = 0 … (2)

x − y = 0 … (3)

To prove:

Sets of given three lines are concurrent.

Explanation:

Now, consider the following determinant:

-b × ab - a × ab – ab × (-a – b) = 0

Hence proved, the given lines are concurrent.

Given:
2x − 5y + 3 = 0 … (1)

5x − 9y + λ = 0 … (2)

x − 2y + 1 = 0 … (3)

To find:

Value of λ.

Concept Used:

Determinant of equation is zero.

Explanation:

It is given that the three lines are concurrent.

∴

⇒ 2(-9 + 2λ) + 5(5 – λ) + 3(-10 + 9) = 0

⇒ -18 + 4λ + 25 – 5λ – 3 = 0

⇒ λ = 4

Hence, λ = 4.

Given:

The given lines can be written as follows:

m₁x – y + c₁ = 0 … (1)

m₂x – y + c₂ = 0 … (2)

m₃x – y + c₃ = 0 … (3)

To find:

Conditions that the straight lines y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ may meet in a point.

Concept Used:

Determinant of equation is zero.

Explanation:

It is given that the three lines are concurrent.

∴

⇒ m₁(-c₃ + c₂) + 1(m₂c₃-m₃c₂) + c₁(-m₂ + m₃) = 0

⇒ m₁(c₂-c₃) + m₂(c₃-c₁) + m₃(c₁-c₂) = 0

Hence, the required condition is m₁(c₂-c₃) + m₂(c₃-c₁) + m₃(c₁-c₂) = 0

Given:

p₁x + q₁y = 1

p₂x + q₂y = 1

p₃x + q₃y = 1

To prove:

The points (p₁, q₁), (p₂, q₂) and (p₃, q₃) are collinear.

Concept Used:

If three lines are concurrent then determinant of equation is zero.

Explanation:

The given lines can be written as follows:

p₁ x + q₁ y – 1 = 0 … (1)

p₂ x + q₂ y – 1 = 0 … (2)

p₃ x + q₃ y – 1 = 0 … (3)

It is given that the three lines are concurrent.

∴

⇒

⇒
Hence proved, This is the condition for the collinearity of the three points, (p₁, q₁), (p₂, q₂) and (p₃, q₃).

Given:

L₁ = (b + c)x + ay + 1 = 0

L₂ = (c + a)x + by + 1 = 0

L₃ = (a + b)x + cy + 1 = 0

To prove:

The straight lines L₁ = (b + c)x + ay + 1 = 0, L₂ = (c + a)x + by + 1 = 0 and L₃ = (a + b)x + cy + 1 = 0 are concurrent.

Concept Used:

If three lines are concurrent then determinant of equation is zero.

Explanation:

The given lines can be written as follows:

(b + c) x + ay + 1 = 0 … (1)

(c + a) x + by + 1 = 0 … (2)

(a + b) x + cy + 1 = 0 … (3)

Consider the following determinant.

Applying the transformation C₁→C₁ + C₂,

⇒

⇒ ⇒

Hence proved, the given lines are concurrent.

Given:

ax + a²y + 1 = 0

bx + b²y + 1 = 0

cx + c²y + 1 = 0

To prove:

At least two of three constant a, b, c are equal.

Concept Used:

If three lines are concurrent then determinant of equation is zero.

Explanation:

The given lines can be written as follows:

ax + a²y + 1 = 0 … (1)

bx + b²y + 1 = 0 … (2)

cx + c²y + 1 = 0 … (3)

The given lines are concurrent.

∴

Applying the transformation R₁→R₁-R₂ and R₂→R₂-R₃:

⇒

⇒ (a – b)(b – c)(c – a) = 0

⇒ a – b = 0 or b – c = 0 or c – a = 0

⇒ a = b or b = c or c = a

Hence proved, atleast two of the constants a,b,c are equal .

Given:

ax + 2y + 1 = 0

bx + 3y + 1 = 0

cx + 4y + 1 = 0

To prove:

The straight lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent.

Concept Used:

If three lines are concurrent then determinant of equation is zero.

Explanation:

The given lines can be written as follows:

ax + 2y + 1 = 0 … (1)

bx + 3y + 1 = 0 … (2)

cx + 4y + 1 = 0 … (3)

Consider the following determinant.

Applying the transformation R₁→R₁-R₂ and R₂→R₂-R₃,

⇒ (-a + b + b – c) = 2b – a – c

Given:
2b = a + c

⇒ a + c –a – c = 0

Hence proved, the given lines are concurrent, provided 2b = a + c.

To prove:

Perpendicular bisectors of the sides of a triangle are concurrent.

Assuming:

ABC be a triangle with vertices A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃).

Let D, E and F be the midpoints of the sides BC, CA and AB, respectively.

Explanation:

Thus, the coordinates of D, E and F are

Let mD, mE and mF be the slopes of AD, BE and CF respectively.

∴ Slope of BC × mD = -1

⇒

⇒ mD

Thus, the equation of AD

⇒

⇒

⇒

Similarly, the respective equations of BE and CF are

Let L₁, L₂ and L₃ represent the lines (1), (2) and (3), respectively.
Adding all the three lines,

We observe:
1 ⋅ L₁ + 1⋅L₂ + 1⋅L₃ = 0

Hence proved, the perpendicular bisectors of the sides of a triangle are concurrent.

Given: equation is parallel to 3x − 4y + 5 = 0 and pass through (2, 3)

To find:

Equation of required line.

Explanation:

The equation of the line parallel to 3x − 4y + 5 = 0 is

3x – 4y + λ = 0,

Where, λ is a constant.

It passes through (2, 3).

∴6 – 12 + λ = 0

⇒ λ = 6

Hence, the required line is 3x − 4y + 6 = 0.

Given: equation is perpendicular to x – 3y + 5 = 0 and passes through (3,-2)

To find:

Equation of required line.

Explanation:

The equation of the line perpendicular to x − 3y + 5 = 0 is

3x + y + λ = 0,

Where λ is a constant.

It passes through (3, − 2).

9 – 2 + λ = 0

⇒ λ = - 7

Substituting λ = − 7 in 3x + y + λ = 0,

Hence, we get 3x + y – 7 = 0, which is the required line.

Given: A (1, 3) and B (3, 1) be the points joining the perpendicular bisector

To find:

The equation of the perpendicular bisector of the line joining the points (1, 3) and (3, 1).

Explanation:

Let C be the midpoint of AB.

∴ coordinates of c

Slope of AB

∴ Slope of the perpendicular bisector of AB = 1

Thus, the equation of the perpendicular bisector of AB is

y – 2 = 1(x – 2)

⇒ x – y = 0

Or, y = x

Hence, the equation is y = x.

Given: The vertices of ∆ABC are A (1, 4), B (− 3, 2) and C (− 5, − 3).

To find:

The equations of the altitudes of a ΔABC whose vertices are A (1, 4), B (-3, 2) and C (-5, -3).

Explanation:

Diagram:

Slope of AB

Slope of BC

Slope of CA

Thus, we have:

Slope of CF = -2

Slope of AD

Slope of BE

Hence,

Equation of CF is: y + 3 = -2(x + 5)

⇒ 2x + y + 13 = 0

Equation of AD is: y – 4(x – 1)

⇒ 2x + 5y – 22 = 0

Equation of BE is: y – 2(x + 3)

⇒ 6x + 7y + 4 = 0

Given: equation is perpendicular to equation and cuts off an intercept of 4 units with the negative direction of y-axis

To find:

The equation of a line which is perpendicular to the line and which cuts off an intercept of 4 units with the negative direction of y-axis.

Explanation:

The line perpendicular to is

It is Given that the line cuts off an intercept of 4 units with the negative direction of the y-axis.

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

∴

⇒

Substituting the value of λ, we get , which is the equation of the required line.

Given: image of (2 1) is (5,2)

To find:

The equation of the mirror.

Explanation:

Let the image of A (2, 1) be B (5, 2).

Also, let M be the midpoint of AB.

∴ Coordinates of M

Diagram:

Let CD be the mirror.

Line AB is perpendicular to the mirror CD.

∴ Slope of AB × Slope of CD = − 1

⇒ Slope of CD = -1

⇒ Slope of CD = -3

Equation of the mirror CD

⇒ 2y – 3 = -6 x + 21

⇒ 6x + 2y – 24 = 0

⇒ 3x + y -12 = 0

Hence, the equation of mirror is 3x + y -12 = 0

Given: equation is perpendicular to lx + my + n = 0 and passing through (α, β)

To find:

The equation of the straight line through the point (α, β) and perpendicular to the line lx + my + n = 0.

Explanation:

The line perpendicular to lx + my + n = 0 is

mx – ly + λ = 0

This line passes through (α, β).

∴ mα-lβ + λ = 0 ⇒ λ = lβ-mα

Substituting the value of λ:

mx – ly + lβ-mα = 0

⇒ mx – α = ly – β

Hence, equation of the required line is mx – α = ly – β

Given: equation is perpendicular to 2x − 3y = 5 and cutting off an intercept 1 on the positive direction of the x-axis.

Explanation:

The line perpendicular to 2x − 3y = 5 is

3x + 2y + λ = 0

It is given that the line 3x + 2y + λ = 0 cuts off an intercept of 1 on the positive direction of the x axis.

This means that the line 3x + 2y + λ = 0 passes through the point (1, 0).

∴ 3 + 0 + λ = 0

⇒ λ = -3

Substituting the value of λ, we get 3x + 2y – 3 = 0,

Hence, equation of the required line.

Given: equation is perpendicular to 5x -2y = 8 and pass through mid-point of the line segment joining (2, 3) and (4, 5).

To find:

The equation of the straight line perpendicular to 5x – 2y = 8 and which passes through the mid-point of the line segment joining (2, 3) and (4, 5).

Explanation:

The line perpendicular to 5x − 2y = 8 is 2x + 5y + λ = 0

Coordinates of the mid points of (2,3) and (4,5)

∴ 6 + 20 + λ = 0

⇒ λ = -26

Substituting the value of λ,

We get 2x + 5y-26 = 0,

Hence, the required equation of line is 2x + 5y-26 = 0.

Given: equation is perpendicular to 3x – 4y + 11 = 0 and has y-intercept equal to

To find:

The equation of the straight line which has y-intercept equal to and is perpendicular to 3x – 4y + 11 = 0.

Explanation:

The line perpendicular to 3x − 4y + 11 = 0 is 4x + 3y + λ = 0

It is given that the line 4x + 3y + λ = 0 has y – intercept equal to

This means that the line passes through

∴ 0 + 4 + λ = 0

⇒ λ = -4

Substituting the value of λ,

We get 4x + 3y – 4 = 0,

Hence, equation of the required line is 4x + 3y – 4 = 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

And, slope of AB

So, the slope of the right bisector of AB is

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

⇒ 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

Given: (2,1) is given point and line mirror is x + y – 5 = 0

To find:

Image of the point with respect to mirror line.

Explanation:

Let the image of A (2, 1) be B (a, b).

Let M be the midpoint of AB.

∴ Coordinates of M are

Diagram:

The point M lies on the line x + y − 5 = 0

⇒ a + b = 7 … (1)

Now, the lines x + y − 5 = 0 and AB are perpendicular.

∴ Slope of AB × Slope of CD = − 1

⇒ a – 2 = b – 1 ……(2)

Adding eq (1) and eq (2):

2a = 8

⇒ a = 4

Now, from equation (1):

4 + b = 7

⇒ b = 3

Hence, the image of the point (2, 1) with respect to the line mirror x + y − 5 = 0 is (4, 3).

Given: image of (2,1) is (5,2)

To find:

The equation of the mirror.

Explanation:

Let the image of A (2, 1) be B (5, 2).

Let M be the midpoint of AB.

Coordinates of M

Diagram:

Let CD be the mirror.

The line AB is perpendicular to the mirror CD.

∴ Slope of AB × Slope of CD = − 1

⇒ Slope of CD = -3

Thus, the equation of the mirror CD is

⇒ 2y – 3 = -6x + 21

⇒ 6x + 2y -24 = 0

⇒ 3x + y – 12 = 0

Hence, the equation of mirror is 3x + y – 12 = 0.

Given: equation parallel to 3x – 4y + 6 = 0 and passing through the middle point of the join of points (2, 3) and (4, -1).

To find:

The equation to the straight line parallel to 3x – 4y + 6 = 0 and passing through the middle point of the join of points (2, 3) and (4, -1).

Explanation:

Let the Given points be A (2, 3) and B (4, − 1). Let M be the midpoint of AB.

∴ Coordinates of M

The equation of the line parallel to 3x − 4y + 6 = 0 is 3x – 4y + λ = 0

This line passes through M (3, 1).

∴ 9 – 4 + λ = 0

⇒ λ = -5

Substituting the value of λ in 3x – 4y + λ = 0, we get 3x – 4y – 5 = 0

Hence, the equation of the required line is 3x – 4y – 5 = 0.

Given: 2x – 3y + 1 = 0,

x + y = 3,

2x – 3y = 2

x + y = 4 are given equation

To prove:

The lines 2x – 3y + 1 = 0, x + y = 3, 2x – 3y = 2 and x + y = 4 form a parallelogram.

Explanation:

The given lines can be written as

… (1)

… (2)

… (3)

… (4)

The slope of lines (1) and (3) is and that of lines (2) and (4) is − 1.

Thus, lines (1) and (3), and (2) and (4) are two pair of parallel lines.

If both pair of opposite sides are parallel then, we can say that it is a parallelogram.

Hence proved, the given lines form a parallelogram.

Given: equation is perpendicular to and it meets the y-axis.

To find:

The equation of a line drawn perpendicular to the line through the point where it meets the y-axis.

Explanation:

Let us find the intersection of the line with y-axis.

At x = 0,

⇒ y = 6

Thus, the given line intersects y-axis at (0, 6).

The line perpendicular to the line is

This line passes through (0, 6).

⇒ λ

Now, substituting the value of λ, we get:

⇒ 2x -3y + 18 = 0

Hence, the equation of the required line is 2x -3y + 18 = 0

Given: perpendicular from the origin and meets at the point (-1, 2)

Explanation:

The given line is y = mx + c which can be written as mx –y + c = 0 … (1)

The slope of the line perpendicular to y = mx + c is

So, the equation of the line with slope and passing through the origin is

x + my = 0 … (2)

Solving eq(1) and eq(2) by cross multiplication, we get

Thus, the point of intersection of the perpendicular from the origin to the line y = mx + c is

It is given that the perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2)

and

⇒ m² + 1 = mc and m² + 1

⇒ mc

⇒ m

Now, substituting the value of m in m² + 1 = mc, we get

⇒ c

Hence, m and c

Given: A (3, 4) and B (− 1, 2) be the given points

To find:

Equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).

Explanation:

Let C be the midpoint of AB.

∴ C

∵ Slope of AB

∴ Slope of the perpendicular bisector of AB = -2

Thus, the equation of the perpendicular bisector of AB is

y-3 = -2(x-1)

⇒ 2x + y – 5 = 0

Hence, the required line is 2x + y – 5 = 0.

Given: A (h,3) and B (4,1) be the points intersect At right angle at the line 7x – 9y – 19 = 0

To find:

The value of h.

Explanation:

The line 7x − 9y − 19 = 0 can be written as

So, the slope of this line is

It is given that the line joining the points A (h,3) and B (4,1) is perpendicular to the line 7x − 9y − 19 = 0.

⇒ 9h – 36 = -14

⇒ 9h = 22

⇒ h

Hence, the value of h is

Given: (3, 8) is given point and line mirror is x + 3y – 7 = 0.

To find:

Image of point with respect to mirror line.

Explanation:

Let the image of A (3,8) be B (a,b).

Also, let M be the midpoint of AB.

∴ Coordinates of M

Diagram:

Point M lies on the line x + 3y = 7

∴

⇒ a + 3b + 13 = 0 … (1)

Lines CD and AB are perpendicular

∴ Slope of AB × Slope of CD = − 1

⇒ b – 8 = 3a – 9

⇒ 3a-b-1 = 0 … (2)

Solving (1) and (2) by cross multiplication, we get:

⇒ a = -1, b = -4

Hence, the image of the point (3, 8) with respect to the line mirror x + 3y = 7 is

(− 1, − 4).

Given: equation is perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.

To find:

The coordinates of the foot of the perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.

Explanation:

Let A (− 1, 3) be the given point.

Also, let M (h, k) be the foot of the perpendicular drawn from A (− 1, 3) to the line 3x − 4y − 16 = 0

Diagram:

Point M (h, k) lies on the line 3x − 4y − 16 = 0

3h − 4k − 16 = 0 … (1)

Lines 3x − 4y − 16 = 0 and AM are perpendicular.

∴

⇒ 4h + 3k – 5 = 0 … (2)

Solving eq (1) and eq (2) by cross multiplication, we get:

Hence, the coordinates of the foot of perpendicular are

Given:

The points (-1, 2) and (5, 4).

To find:

The projection of the point (1, 0) on the line joining the points (-1, 2) and (5, 4).

Explanation:

Let A (− 1, 2) be the given point whose projection is to be evaluated and C (− 1, 2) and D (5, 4) be the other two points.

Also, let M (h, k) be the foot of the perpendicular drawn from A (− 1, 2) to the line joining the points C (− 1, 2) and D (5, 4).

Diagram:

Clearly, the slope of CD and MD are equal.

⇒ h – 3k + 7 = 0…….(1)

The lines segments AM and CD are perpendicular.

⇒ 3h + k – 3 = 0 …….(2)

Solving (1) and (2) by cross multiplication, we get:

Hence, the projection of the point (1, 0) on the line joining the points (-1, 2) and (5, 4) is

Given:

Line and distance of 3 units from the origin.

To find:

The equation of a line perpendicular to the line and at a distance of 3 units from the origin.

Explanation:

The line perpendicular to is

It is given that the line is at a distance of 3 units from the origin.

⇒ λ = ± 6

Substituting the value of λ,

We get ,

Hence, equation of the required line is

Given:

Line 2x + 3y = 12 meets the x-axis at A and y-axis at B

To find:

The area of figure OCEB.

Explanation:

The given line is 2x + 3y = 12, which can be written as

……(1)

So, the coordinates of the points A and B are (6, 0) and (0, 4), respectively.

Diagram:

The equation of the line perpendicular to line (1) is

This line passes through the point (5, 5).

Now, substituting the value of λ in , we get:

⇒ …….(2)

Thus, the coordinates of intersection of line (1) with the x-axis is C

To find the coordinates of E, let us write down equations (1) and (2) in the following manner:

2x + 3y – 12 = 0 … (3)

3x – 2y – 5 = 0 … (4)

Solving (3) and (4) by cross multiplication, we get:

⇒ x = 3, y = 2

Thus, the coordinates of E are (3, 2).

From the figure,

EC

EA

Now,

Area(OCEB) = Area(∆OAB) –Area(∆CAE)

⇒ Area(OCEB)

sq units

Hence, area of figure OCEB is sq units

To find:

The equation of the straight line which cuts off intercepts on x-axis twice that on y-axis and is at a unit distance from the origin.

Assuming:

Intercepts on x-axis and y-axis be 2a and a, respectively.

Explanation:

So, the equation of the line with intercepts 2a on x-axis and a on y-axis be

⇒ x + 2y = 2a … (1)

Let us change equation (1) into normal form.

Thus, the length of the perpendicular from the origin to the line (1) is

Given:

P = 1

Required equation of the line:

x + 2y

⇒ x + 2y + = 0

Hence, equation of required line is x + 2y + = 0.

Given:

Sides AB and AC of a triangle ABC are x – y + 5 = 0 and x + 2y = 0.

To find:

The equation of the line BC.

Explanation:

Diagram:

Let the perpendicular bisectors x − y + 5 = 0 and x + 2y = 0 of the sides AB and AC intersect at D and E, respectively.

Let (x₁,y₁) and (x₂,y₂) be the coordinates of points B and C.

Coordinates of D

And coordinates of E

Point D lies on the line x − y + 5 = 0

⇒ x₁ – y₁ + 13 = 0 … (1)

Point E lies on the line x + 2y = 0

⇒ x₂ + 2y₂ – 3 = 0 … (2)

Side AB is perpendicular to the line x − y + 5 = 0

⇒ x₁ + y₁ + 1 … (3)

Similarly, side AC is perpendicular to the line x + 2y = 0

⇒ 2x₂ - y₂ - 4 = 0… (4)

Now, solving eq (1) and eq (3) by cross multiplication, we get:

⇒ x₁ = -7, y₁ = 6

Thus, the coordinates of B are (-7, 6)

Similarly, solving (2) and (4) by cross multiplication, we get:

⇒

Thus, coordinates of C are

Therefore, equation of line BC is

⇒

⇒ 14x + 23 y – 40 = 0

Hence, the equation of line BC is 14x + 23 y – 40 = 0

Given:

The equations of the lines are

3x + y + 12 = 0 … (1)

x + 2y − 1 = 0 … (2)

To find:

Angles between two lines.

Explanation:

Let m₁ and m₂ be the slopes of these lines.

m₁ = -3, m₂

Let θ be the angle between the lines.
Then,

⇒

Hence, the acute angle between the lines is 45°

Given:

The equations of the lines are

3x − y + 5 = 0 … (1)

x − 3y + 1 = 0 … (2)

To find:

Angles between two lines.

Explanation:

Let m₁ and m2 be the slopes of these lines.

m₁ = 3, m₂

Let θ be the angle between the lines.
Then,

⇒

Hence, the acute angle between the lines is

Given:

The equations of the lines are

3x + 4y − 7 = 0 … (1)

4x − 3y + 5 = 0 … (2)

To find:

Angles between two lines.

Explanation:

Let m₁ and m₂ be the slopes of these lines.

m₁m₂

∵m₁m₂

Hence, the given lines are perpendicular.
Therefore, the angle between them is 90°.

Given:

The equations of the lines are

x − 4y = 3 … (1)

6x − y = 11 … (2)

To find:

Angles between two lines.

Explanation:

Let m₁ and m₂ be the slopes of these lines.

m₁m₂ = 6

Let θ be the angle between the lines.
Then,

⇒

Hence, the acute angle between the lines is

Given:

The equations of the lines are

(m² − mn) y = (mn + n²) x + n³ … (1)

(mn + m²) y = (mn − n²) x + m³ … (2)

To find:

Angles between two lines.

Explanation:

Let m₁ and m₂ be the slopes of these lines.

∴

Let θ be the angle between the lines.

Then,

⇒

⇒

Then,

⇒

⇒

Hence, the acute angle between the lines is

Given:

The equations of the lines are

2x − y + 3 = 0 … (1)

x + y + 2 = 0 … (2)

To find:

Angles between two lines.

Explanation:

Let m₁ and m₂ be the slopes of these lines.

m₁ = 2, m₂ = -1

Let θ be the angle between the lines.
Then,

⇒

Hence, the acute angle between the lines is

To prove:

The points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram

To find:

The angle between diagonals of parallelogram.

Assuming:

A(2, − 1), B(0, 2), C(2, 3) and D(4, 0) be the vertices.

Explanation:

Slope of AB

Slope of BC

Slope of CD

Slope of DA

Thus, AB is parallel to CD and BC is parallel to DA.

Therefore, the given points are the vertices of a parallelogram.

Now, let us find the angle between the diagonals AC and BD.

Let m₁ and m₂ be the slopes of AC and BD, respectively.

∴

∴

Thus, the diagonal AC is parallel to the y-axis.

∴∠ODB

In triangle MND,

∠DMN

Hence proved, the acute angle between the diagonal is .

Given:

Points (2, 0), (0, 3) and the line x + y = 1.

Assuming:

Let A (2, 0), B (0, 3) be the given points.

To find:

Angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Explanation:

Slope of AB = m₁

Slope of the line x + y = 1 is -1

∴ m₂ = -1

Let θ be the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1

∴

⇒

Hence, the acute angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1 is .

To prove:

and

Assuming:

A (x₁, y₁) and B (x₂, y₂) be the given points and O be the origin.

Explanation:

Slope of OA = m₁ = y_1x1

Slope of OB = m₂ = y_2x2

It is given that θ is the angle between lines OA and OB.

∴

⇒

Now,

As we know that

∴

⇒

⇒

Hence proved.

Given:

The given lines are

(a + b) x + (a − b) y = 2ab … (1)

(a − b) x + (a + b) y = 2ab … (2)

x + y = 0 … (3)

To prove:

The straight lines (a + b)x + (a – b)y = 2ab, (a – b)x + (a + b)y = 2ab and x + y = 0 form an isosceles triangle whose vertical angle is

Assuming:

Let m₁, m₂ and m₃ be the slopes of the lines (1), (2) and (3), respectively.

Explanation:

Now,

Slope of the first line = m₁

Slope of the second line = m₂

Slope of the third line = m₃ = -1

Let θ₁ be the angle between lines (1) and (2), θ₂ be the angle between lines (2) and (3) and θ₃ be the angle between lines (1) and (3).

∴

⇒

⇒

∴

⇒

⇒

∴

⇒

⇒

Here,
θ₂ = θ₃ and

Hence proved, the given lines form an isosceles triangle whose vertical angle is

Given:

x = a

by + c = 0

To find:

Angle between the lines x = a and by + c = 0.

Concept Used:

Angle between two lines.

Explanation:

The given lines can be written as

x = a … (1)

y = -cb … (2)

Hence, Lines (1) and (2) are parallel to the y-axis and x-axis, respectively. Thus, they intersect at right angle, i.e. at 90°.

Given:

Intercepts 3, 4 and 1, 8 on the axes

To find:

Tangent of the angle between the lines

Concept Used:

Angle between two lines.

Explanation:

The respective equations of the lines having intercepts 3, 4 and 1, 8 on the axes are

… (1)

… (2)

Assuming:

m₁ and m₂ be the slope of the lines (1) and (2), respectively.

∴m₁, m₂ = -8

Let θ be the angle between the lines (1) and (2).

∴

⇒

Hence, the tangent of the angles between the lines is .

Given:

Line a²x + ay + 1 = 0 is perpendicular to the line x – ay = 1

To prove:

The line a²x + ay + 1 = 0 is perpendicular to the line x – ay = 1 for all non-zero real values of a.

Concept Used:

Product of slope of perpendicular line is -1.

Explanation:

The given lines are

a²x + ay + 1 = 0 … (1)

x − ay = 1 … (2)

Let m₁ and m₂ be the slopes of the lines (1) and (2).

m₁m₂ = -

Hence proved, line a²x + ay + 1 = 0 is perpendicular to the line x − ay = 1 for all non-zero real values of a.

Given:

……(i)

……(ii)

To prove:

The tangent of an angle between the lines and is

Concept Used:

Angle between two lines.

Assuming:

m₁ and m₂ be the slope of the lines (1) and (2), respectively.

Explanation:

∴m₁, m₂

Let θ be the angle between the lines (1) and (2).

∴

Hence proved, the tangent of the angles between the lines is .

Given: x – 5y + 6 = 0, x – 3y + 2 = 0 and x – 2y – 3 = 0 forming a triangle and point P(α², α) lies inside or on the triangle

To find: value of α

Explanation:

Let ABC be the triangle of sides AB, BC and CA whose equations are x − 5y + 6 = 0, x − 3y + 2 = 0 and x − 2y − 3 = 0, respectively.
On solving the equations,

We get A (9, 3), B (4, 2) and C (13, 5) as the coordinates of the vertices.

Diagram:

It is given that point P (α², α) lies either inside or on the triangle. The three conditions are given below.

(i) A and P must lie on the same side of BC.

(ii) B and P must lie on the same side of AC.

(iii) C and P must lie on the same side of AB.

If A and P lie on the same side of BC, then

(9 – 9 + 2)(α² – 3α + 2) ≥0

⇒ (α – 2)(α – 1) ≥ 0

⇒ α ∈ (- ∞, 1 ] ∪ [ 2, ∞) … (1)

If B and P lie on the same side of AC, then

(4 – 4 – 3) (α² – 2α – 3) ≥ 0

⇒ (α – 3)(α + 1) ≤ 0

⇒ α ∈ [- 1, 3] … (2)

If C and P lie on the same side of AB, then

(13 – 25 + 6)(α² – 5α + 6) ≥0

⇒ (α – 3)(α – 2) ≤ 0

⇒ α ∈ [ 2, 3] … (3)

From (1), (2) and (3), we get:

α∈ [ 2, 3]

Hence, α∈ [2, 3]

Given:

x + y – 4 = 0, 3x – 7y – 8 = 0 and 4x – y – 31 = 0 forming a triangle and point (a, 2)is an interior point of the triangle

To find:

Value of a

Explanation:

Let ABC be the triangle of sides AB, BC and CA whose equations are x + y − 4 = 0, 3x − 7y − 8 = 0 and 4x − y − 31 = 0, respectively.

On solving them, we get A (7, - 3), B and C as the coordinates of the vertices.
Let P (a, 2) be the given point.

Diagram:

It is given that point P (a, 2) lies inside the triangle. So, we have the following:

(i) A and P must lie on the same side of BC.

(ii) B and P must lie on the same side of AC.

(iii) C and P must lie on the same side of AB.

Thus, if A and P lie on the same side of BC, then

21 + 21 – 8 – 3a – 14 – 8 > 0

⇒ a > … (1)

If B and P lie on the same side of AC, then

⇒ a < … (2)

If C and P lie on the same side of AB, then

⇒

⇒ a > 2 … (3)

From (1), (2) and (3), we get:

A ∈

Hence, A ∈

Given:

x + y – 4 = 0, 3x – 7y + 8 = 0, 4x – y – 31 = 0 forming a triangle and point (-3, 2)

To find:

Point (-3, 2) lies inside or outside the triangle

Explanation:

Let ABC be the triangle of sides AB, BC and CA, whose equations x + y − 4 = 0, 3x − 7y + 8 = 0 and 4x − y − 31 = 0, respectively.

On solving them, we get A (7, - 3), B (2, 2) and C (9, 5) as the coordinates of the vertices.

Let P (− 3, 2) be the given point.

Diagram:

The given point P (− 3, 2) will lie inside the triangle ABC, if

(i) A and P lies on the same side of BC

(ii) B and P lies on the same side of AC

(iii) C and P lies on the same side of AB

Thus, if A and P lie on the same side of BC, then

21 + 21 + 8 – 9 – 14 + 8 > 0

⇒ 50 × - 15 > 0

⇒ - 750 > 0,

Which is false

Hence, the point (− 3, 2) lies outside triangle ABC.

Given:

Line 3x – 5y + 7 = 0

Concept Used:

Distance of a point from a line.

To find:

The distance of the point (4, 5) from the straight line 3x – 5y + 7 = 0.

Explanation:

Comparing ax + by + c = 0 and 3x − 5y + 7 = 0, we get:

a = 3, b = − 5 and c = 7

So, the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0 is

d

⇒ d

Hence, the required distance is

Given: points (cosθ, sinθ) and (cosϕ, sinϕ) from the origin.

To find:

The perpendicular distance of the line joining the points (cosθ, sinθ) and (cosϕ, sinϕ) from the origin

Concept Used:

Distance of a point from a line.

Explanation:

The equation of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) is given below:

⇒

⇒

Let d be the perpendicular distance from the origin to the line

-sinθ2 + cosϕ-cosθ2

⇒ d

⇒ d

⇒ d

⇒ d

⇒ d

⇒ d

⇒ d

Hence, the required distance is

Given:

Coordinates are (a cos α, a sin α) and (a cos β, a sin β).

To find:

The length of the perpendicular from the origin to the straight line joining the two points whose coordinates are (a cos α, a sin α) and (a cos β, a sin β).

Concept Used:

Distance of a point from a line.

Explanation:

Equation of the line passing through (acosα, asinα) and (acosβ, asinβ) is

The distance of the line from the origin is

Hence, the required distance is

Given:

Lines 24x + 7y = 20 and 4x – 3y – 2 = 0

To prove:

The perpendicular let fall from any point on the straight line 2x + 11y – 5 = 0 upon the two straight lines 24x + 7y = 20 and 4x – 3y – 2 = 0 are equal to each other.

Concept Used:

Distance of a point from a line.

Assuming:

P(a, b) be any point on 2x + 11y − 5 = 0

Explanation:

∴ 2a + 11b − 5 = 0

⇒ b ……..(i)

Let d₁ and d₂ be the perpendicular distances from point P
on the lines 24x + 7y = 20 and 4x − 3y − 2 = 0, respectively.

d₁

⇒ d₁

From (1)

⇒ d₁

Similarly,

d₂

⇒ d₂

From (1)

⇒ d₂

∴ d₁ = d₂

Hence proved.

Given:

Lines 2x + 3y = 21 and 3x – 4y + 11 = 0

To find:

The distance of the point of intersection of the lines 2x + 3y = 21 and 3x – 4y + 11 = 0 from the line 8x + 6y + 5 = 0.

Concept Used:

Distance of a point from a line.

Explanation:

Solving the lines 2x + 3y = 21 and 3x − 4y + 11 = 0 we get:

⇒ x = 3, y = 5

So, the point of intersection of 2x + 3y = 21 and 3x − 4y + 11 = 0 is (3, 5).

Now, the perpendicular distance d of the line 8x + 6y + 5 = 0 from the point (3, 5) is
d

Hence, distance is .

Given:

Lines 2x – 3y + 14 = 0 and 5x + 4y – 7 = 0.

To find:

The length of the perpendicular from the point (4, -7) to the line joining the origin and the point of intersection of the lines 2x – 3y + 14 = 0 and 5x + 4y – 7 = 0.

Concept Used:

Distance of a point from a line.

Explanation:

Solving the lines 2x − 3y + 14 = 0 and 5x + 4y − 7 = 0 we get:

⇒ x, y

So, the point of intersection of 2x − 3y + 14 = 0 and 5x + 4y − 7 = 0 is

The equation of the line passing through the origin and the point is

⇒ y

⇒ y

⇒ 12x + 5y = 0

Let d be the perpendicular distance of the line 12x + 5y = 0 from the point (4, − 7)

∴d

Hence, Length of perpendicular is 1.

Given:

The points on x-axis whose perpendicular distance from the straight line is a

To find:

Points on x-axis

Concept Used:

Distance of a point from a line.

Explanation:

Let (t, 0) be a point on the x-axis.

It is given that the perpendicular distance of the line from a point is a.

∴

⇒

⇒

⇒

⇒

⇒

⇒ t

Hence, the required points on the x-axis are and

Given:

and points

To prove:

The product of perpendicular on the line from the points is b².

Concept Used:

Distance of a point from a line.

Explanation:

Let d₁ and d₂ be the perpendicular distances of line from points and respectively.

∴d₁

Similarly,

d₁ = -

Now,

d₁d₂

⇒ d₁d₂

⇒ d₁d₂

⇒ d₁d₂

Hence proved.

Given:

Line

To find:

The perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line .

Concept Used:

Distance of a point from a line.

Explanation:

The equation of the line perpendicular to is
This line passes through (1, 2).

∴ + 2 + λ = 0

⇒ λ

Substituting the value of λ, we get

Let d be the perpendicular distance from the origin to the line

d

Hence, the required perpendicular distance is

Given:

Lines x + 2y = 5 and x – 3y = 7, slope = 5.

To find:

The distance of the point (1, 2) from the straight line with slope 5 and passing through the point of intersection of x + 2y = 5 and x – 3y = 7.

Concept Used:

Distance of a point from a line.

Explanation:

To find the point intersection of the lines x + 2y = 5 and x − 3y = 7, let us solve them.

⇒ x y

So, the equation of the line passing through with slope 5 is

⇒ 5y + 2 = 25x – 145

⇒ 25x – 5y – 147 = 0

Let d be the perpendicular distance from the point (1, 2) to the line 25x – 5y – 147 = 0

∴d

Hence, the required perpendicular distance is

Given:

Distance from the line is 4.

To find:

Points on y-axis

Concept Used:

Distance of a point from a line.

Explanation:

Let (0, t) be a point on the y-axis.

It is given that the perpendicular distance of the line from the point (0, t) is 4 units.

∴

⇒

⇒

⇒

⇒ t

⇒ t

⇒ t

Hence, the required points on the y-axis are and .

Given:

A(2, 3), B(4, -1) and C(1, 2).

To find:

The equation and the length of the altitude from the vertex A.

Concept Used:

Distance of a point from a line.

Explanation:

Equation of side BC:

⇒ x + y – 3 = 0

The equation of the altitude that is perpendicular to x + y – 3 = 0 is x – y + λ = 0.

Line x – y + λ = 0 passes through (2, 3).

∴ 2 – 3 + λ = 0

⇒ λ = 1

Thus, the equation of the altitude from the vertex A (2, 3) is x – y + 1 = 0.

Let d be the length of the altitude from A (2, 3).

d

⇒ d

Hence, the required distance is.

Given:

Two lines 3x – 2y = 5 and 3x + 2y = 5

To prove:

The path of a moving point such that its distances from two lines 3x – 2y = 5 and 3x + 2y = 5 are equal is a straight line

Concept Used:

Distance of a point from a line.

Explanation:

Let P(h, k) be the moving point such that it is equidistant from the lines 3x − 2y = 5 and 3x + 2y = 5

⇒ |3h – 2k – 5| = |3h + 2k – 5|

⇒ 3h – 2k – 5 = ±(3h + 2k – 5)

⇒ 3h – 2k – 5 = 3h + 2k – 5 and 3h – 2k – 5 = -3h + 2k – 5

⇒ k = 0 and 3h = 5

Hence proved, the path of the moving points are 3x = 5 or y = 0. These are straight lines.

Given:

Sum of perpendicular distances of a variable point P(x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10

To prove:

P must move on a line.

Concept Used:

Distance of a point from a line.

Explanation:

It is given that the sum of perpendicular distances of a variable point P (x, y) from the lines x + y − 5 = 0 and 3x − 2y + 7 = 0 is always 10

∴

⇒

It is a straight line.

Hence proved.

Given:

Line ax – by + c = 0 and point (1, 1)

To prove:

Concept Used:

Distance of a point from a line.

Explanation:

The distance of the point (1, 1) from the straight line ax − by + c = 0 is 1

∴ 1

⇒ a² + b² + c² – 2ab + 2ac – 2bc = a² + b²

⇒ ab + bc – ac

Dividing both the sides by abc, we get:

Hence proved.

Given: The parallel lines are

4x − 3y − 9 = 0 … (1)

4x − 3y − 24 = 0 … (2)

To find:

Distance between the givens parallel lines

Explanation:

Let d be the distance between the given lines.

Hence, distance between givens parallel line is

Given: The parallel lines are

8x + 15y − 34 = 0 … (1)

8x + 15y + 31 = 0 … (2)

To find:

Distance between the givens parallel lines

Explanation:

Let d be the distance between the given lines.

Hence, distance between givens parallel line is

Given: The parallel lines are

y = mx + c and y = mx + d

To find:

Distance between the givens parallel lines

Explanation:

The parallel lines can be written as

mx − y + c = 0 … (1)

mx − y + d = 0 … (2)

Let d be the distance between the given lines.

Hence, distance between givens parallel line is

Given: The parallel lines are

4x + 3y – 11 = 0 and 8x + 6y = 15

To find:

Distance between the givens parallel lines

Explanation:

The given parallel lines can be written as

4x + 3y − 11 = 0 … (1)

… (2)

Let d be the distance between the given lines.

Hence, distance between givens parallel line is

Given: Two side of square are 5x – 12y – 65 = 0 and 5x – 12y + 26 = 0

To find: area of the square

Explanation:

The sides of a square are

5x − 12y − 65 = 0 … (1)

5x − 12y + 26 = 0 … (2)

We observe that lines (1) and (2) are parallel.

So, the distance between them will give the length of the side of the square.

Let d be the distance between the given lines.

∴ Area of the square = 7²
= 49 square units

Given: equation is parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1)

To find: equation of two straight lines

Explanation:

The equation of given line is

x + 7y + 2 = 0 … (1)

The equation of a line parallel to line x + 7y + 2 = 0 is given below:

X + 7y + λ = 0 … (2)

The line x + 7y + λ = 0 is at a unit distance from the point (1, − 1).

∴ 1 = 1 – 7 + λ1 + 49

Hence, the required lines:

and

Given: lines A 2x + 3y = 19 and B 2x + 3y + 7 = 0 also a line C 2x + 3y = 6.

To prove:

Line A and B are equidistant from the line C

Proof:

Let d1 be the distance between lines 2x + 3y = 19 and 2x + 3y = 6,

While d2 is the distance between lines 2x + 3y + 7 = 0 and 2x + 3y = 6

Hence proved, the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6

Given:

9x + 6y – 7 = 0 and 3x + 2y + 6 = 0 are parallel lines

To find:

The equation of the line mid-way between the givens line

Explanation:

The given equations of the lines can be written as:

… (1)

3x + 2y + 6 = 0 … (2)

Let the equation of the line midway between the parallel lines (1) and (2) be

3x + 2y + λ = 0 … (3)

The distance between (1) and (3) and the distance between (2) and (3) are equal.

Equation of the required line:

⇒ 18x + 12y + 11 = 0

Hence, equation of required line is 18x + 12y + 11 = 0.

Given:

Lines A: 3x + 4y + 5 = 0 and B: 3x + 4y – 5 = 0

And also C: 3x + 4y + 2 = 0

To find:

Ratio in which line C divides the distance between the lines A and B

Explanation:

The distance between two parallel line ax + by + c1 = 0 and ax + by + c2 = 0 is

Therefore the distance between two parallel line 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is

The distance between the line 3x + 4y + 2 = 0 and 3x + 4y – 5 = 0 is

Hence, the required ratio

Given:

The given lines are

a₁x + b₁y + c₁ = 0 … (1)

a₁x + b₁y + d₁ = 0 … (2)

a₂x + b₂y + c₂ = 0 … (3)

a₂x + b₂y + d₂ = 0 … (4)

To prove:

The area of the parallelogram formed by the lines

a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0, a₂x + b₂y + d₂ = 0 is sq. units.

Explanation:

The area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0 and a₂x + b₂y + d₂ = 0 is given below:

Area

∵a₁b₂ – a₂b₁

∴ Area

If the given parallelogram is a rhombus, then the distance between the pair of parallel lines are equal.

∴

Hence proved.

Given:

The given lines are

3x − 4y + a = 0 … (1)

3x − 4y + 3a = 0 … (2)

4x − 3y − a = 0 … (3)

4x − 3y − 2a = 0 … (4)

To prove:

The area of the parallelogram formed by the lines 3x – 4y + a = 0, 3x – 4y + 3a = 0, 4x – 3y – a = 0 and 4x – 3y – 2a = 0 is sq. units.

Explanation:

From Above solution, We know that

Area of the parallelogram

⇒ Area of the parallelogramsquare units

Hence proved.

Given:

The given lines are

lx + my + n = 0 … (1)

mx + ly + n’ = 0 … (2)

lx + my + n’ = 0 … (3)

mx + ly + n = 0 … (4)

To prove:

The diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n’ = 0, mx + ly + n = 0 and mx + ly + n’ = 0 include an angle.

Explanation:

Solving (1) and (2), we get,

B

Solving (2) and (3), we get,

C

Solving (3) and (4), we get,

D

Solving (1) and (4), we get,

A

Let m₁ and m₂ be the slope of AC and BD.

∴m₁m₂ = -1

Hence proved, diagonals of the parallelogram intersect at an angle

Given:

Equation passes through (0, 0) and make an angle of 45° with the line

To find:

Equation of given line

Explanation:

We know that, the equations of two lines passing through a point x1,y1 and making an angle α with the given line y = mx + c are

Here,
x₁ = 0, y₁ = 0, α = 45° and m =

So, the equations of the required lines are

Hence, Equation of given line is

Given:

Equation passes through (0,0) and make an angle of 75° with the line

To find:

Equation of given line

Explanation:

We know that the equations of two lines passing through a point x1,y1 and making an angle α with the given line y = mx + c are

Here,
Equation of the given line is,

Comparing this equation with y = mx + c

We get,

∴ x₁ = 0, y₁ = 0, α = 75°, m

and tan75∘ = 2 +

So, the equations of the required lines are