Theorem – 1 (Vector Equation of Line in Space):

The vector equation of a straight line passing through a fixed point with position vector a  and parallel to a given vector b is r = a + λ b. Where λ is scalar and called the parameter.

Notes:

• In the above equation r is a position vector of any point P(x, y, z) on the line, then r = x   + y  + z  
• The position vector of any point on the line is taken as a + λ b. This form of the equation is called the vector form.
• If the line passes through the origin its vector equation is r = λ b
• The vector equation of a line passing through a fixed point with position vector a and parallel to a given vector b is also given as

(r – a) × b

Theorem – 2 (Cartesian Equation of Line in Space):

The cartesian equation of a straight line passing through a fixed point P(x₁, y₁, z₁) and having direction ratios (d.r.s) proportional to a, b, c respectively is given by 

Notes:

• If then x = aλ + x₁, y = bλ + y₁, and z = cλ + z₁. These equations are called the parametric equations of the line.
• The coordinates of any point on the line are (aλ + x₁, bλ + y₁, cλ + z₁).
• If the line passes through the origin its equation is

• Since the direction cosines (d.c.s) of a line are direction ratios (d.r.s) of the line, the equation of the line passing through a fixed point P(x₁, y₁, z₁) and having direction cosines (d.c.s) l, m, n respectively is given by

• The x-axis, y-axis, and z-axis pass through origin.
• d.r.s of x-axis are 1, 0, 0. Hence its equation is y = 0 and z = 0.
• d.r.s ofy-axiss are 0, 1, 0. Hence its equation is x = 0 and z = 0.
• d.r.s of z – axis are 0, 0, 1. Hence its equation is x = 0 and y = 0.

Theorem – 3:

The vector equation of a straight line passing through two fixed points with position vector a and b  is

r = a + λ( b – a)

Where λ is scalar and called the parameter.

Theorem – 4:

The cartesian equation of a straight line passing through two fixed points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is given by

To find Direction Ratios and Direction Cosines of Line in Space:

Algorithm:

• Write the equation in standard form
• Make sure there are no coefficients for the x, y, and z terms. If coefficients are present divide the numerator and denominator by the coefficient.
• Do simplification if any
• Then the denominators indicate the direction ratios.
• Using direction ratios, find direction cosines

Example 01:

• Find the direction cosines of the line 
• Solution:

The equation of the line is

Writing in a standard form

The d.r.s of lines are 2, 3/2, 0 i.e. 4, 3, 0 ≡ a, b, c
Now d.c. s of the line are

Ans: d.c.s of the line are 4/5, 3/5, 0

Example 02:

• Find the direction cosines of the line 
• Solution:

The equation of the line is 

Writing in a standard form

The d.r.s of lines are -2, 6, -3 ≡ a, b, c
Now d.c. s of the line are

Ans: d.c.s of the line are -2/7, 6/7, -3/7

Example 03:

• Find the direction cosines of the line 
• Solution:

The equation of the line is

Writing in a standard form

The d.r.s of lines are 2, 3, 0 ≡ a, b, c
Now d.c. s of the line are

Ans: d.c.s of the line are 2/√13, 3/√13, 0.

Example 04:

• Find the direction cosines of the line 
• Solution:

The equation of the line is

Writing in a standard form

The d.r.s of lines are 2, 4, 3 ≡ a, b, c
Now d.c. s of the line are

Ans: d.c.s of the line are -2/√17, 2/√17, -3/√17.

Conversion of Vector Equation into Cartesian Equation of Line in Space:

Algorithm:

1. Equate the vector form r = a + λ b to r = x   + y  + z 
2. Open the brackets of R.H.S.
3. Group the terms of , , and 
4. Equate corresponding terms on both the sides
5. Find three distinct equations for λ.
6. Equate the three equations

Example 05:

• Find the cartesian equations of a line whose vector equation is  r = (2   –   +  4) + λ(   +   –  2)
• Solution:

The vector equation of the line is   

r = (2   –   +  4) + λ(   +   –  2)

Where, r = x   + y  + z 

∴  x   + y  + z  = (2   –   +  4) + λ(   +   –  2)

∴  x   + y  + z   = 2   –   +  4 + λ  + λ –  2λ

∴  x   + y  + z   = (2 + λ)   + (-1 + λ)  +  (4 – 2 λ)

∴ x = 2 + λ  ⇒  λ = x – 2 …………. (1)

∴ y = -1 + λ  ⇒  λ = y + 1 …………. (2)

∴ z = 4 – 2λ  ⇒  λ = (z – 4)/(-2)  …………. (3)

From equations (1), (2) and (3)

x – 2 = y + 1 = (z – 4)/(-2)

Thus the cartesian equations of lines are

Conversion of Cartesian Equation into Vector Equation of Line in Space:

Algorithm (Method – I):

1. Write given the cartesian equation in standard form. 
2. Then write the position vector of the point through which the line is passing. a = x₁  + y₁ + z₁
3. The direction ratios of the line are a, b, and c. Write the direction vector,  b = a   + b  + c 
4. Write the vector form of the equation as r = a + λ b. Where λ ∈ R, and is a scalar/parameter
5. Thus vector equation of line is r = (x₁ + y₁ + z₁)+ λ (a  + b  + c )

Algorithm (Method – II):

1. Let 
2. Find x = aλ + x₁, y = bλ + y₁, and z = cλ + z₁,
3. Substitute values of a, y, and z in the equation r = x   + y  + z  
4. Group terms on R.H.S without λ and with λ.
5. Get the vector equation in the format r = (x₁ + y₁ + z₁)+ λ (a  + b  + c )

Example 06:

• Find the vector equation of a line whose cartesian equation is  6x – 2 = 3y + 1 = 2z – 2
• Solution (Method – I):

The cartesian equation of the line is

Thus the line passes through the point (1/3, -1/3, 2)
The position vector of this point is  a =(1/3)  – (1/3) + 2

The d.r. s of the line are i.e. 1, 2, 3 ≡ a, b, c
Hence, the direction vector of the line is b =   + 2 + 3

Now, the vector form of the equation of the line is given by

 r = a + λ b

Where λ ∈ R, and is a scalar/parameter

Example 07:

• Find the vector equation of a line whose cartesian equation is 
• Solution (Method – II):

The equation of the line is 

Let,   = λ
Thus x = 3λ – 5, y = 5λ – 4 and z = 6λ – 5
Now, r = x   + y  + z 

 r = (3λ – 5)   + (5λ – 4)  + (6λ – 5) 

 r = 3λ   – 5    + 5λ  – 4  + 6λ  – 5 

 r =  – 5    – 4  – 5  + 3λ   +  5λ   + 6λ 

 r =  (- 5    – 4  – 5  ) + λ(3 +  5  + 6  )

Where λ ∈ R, and is a scalar/parameter

Example 08:

• Find the vector equation of a line whose cartesian equation is 
• Solution:

The equation of the line is 

Thus the line passes through the point (6, -4, 5)
The position vector of this point is  a = 6  – 4 + 5

The d.r. s of the line are 2, 7, 3≡ a, b, c
Hence, the direction vector of the line is  b = 2  + 7 + 3

Now, the vector form of the equation of the line is given by

 r = a + λ b

 r = (6  – 4 + 5) + λ (2  + 7 + 3)

Where λ ∈ R, and is a scalar/parameter

Example 09:

• Find the vector equation of a line whose cartesian equation is 
• Solution:

The cartesian equation of the line is 

Thus the line passes through the point (5, -4, 6)
The position vector of this point is  a = 5  – 4 + 6

The d.r. s of the line are 3, 7, 2 ≡ a, b, c
Hence, the direction vector of the line is b = 3  + 7 + 2

Now, the vector form of the equation of the line is given by

 r = a + λ b

 r = (5  – 4 + 6) + λ (3  + 7 + 2)

Where λ ∈ R, and is a scalar/parameter

Example 10:

• Find the vector equation of a line whose cartesian equation is x = ay + b and z = cy + d . Find its direction ratios.
• Solution:

The cartesian equation of the line is

Thus the line passes through the point(b, 0, d)
The position vector of this point is  a = b  + 0 + d = b  + d

The d.r. s of the line are a, 1, c ≡ a, b, c
Hence, the direction vector of the line is  b = a  +  + c

Now, the vector form of the equation of the line is given by

r = a + λ b

 r = (b  + d) + λ (a  +  + c)

Where λ ∈ R, and is a scalar/parameter

Example 11:

• Find the vector equation of a line whose cartesian equation is 3x + 1 = 6y – 2 = 1- z . Find the fixed point through which it passes and its. d.r.s
• Solution:

The equation of the line is 3x + 1 = 6y – 2 = 1- z

Thus the line passes through the point ( -1/3, 1/3, 1)
The position vector of this point is a = (-1/3)  + (1/3) +

The d.r. s of the line are i.e. 1/3, 1/6, -1 ≡ a, b, c
Hence, the direction vector of the line is B = (1/3)  + (1/6) –

Now, the vector form of the equation of the line is given by

r = a + λ b

Where λ ∈ R, and is a scalar/parameter

Example 12:

• Find the vector equation of a line whose cartesian equation is 2x – 2 = 3y + 1 = 6z – 2. Find the fixed point through which it passes and its. d.r.s.
• Solution:

The equation of the line is 2x – 2 = 3y + 1 = 6z – 2

Thus the line passes through the point (1, – 1/3, 1/3)
The position vector of this point is a =    – (1/3) + (1/3)

The d.r. s of the line are i.e. 3, 2, 1≡ a, b, c
Hence, the direction vector of the line is b = 3   + 2 +

Now, the vector form of the equation of the line is given by

r = a + λ b

Where λ ∈ R, and is a scalar/parameter

Example 13:

• Find the vector equation of a line whose cartesian equation is 3x – 1 = 6y + 2 = 1 – z
• Solution:

The cartesian equation of the line is 3x – 1 = 6y + 2 = 1 – z

Thus the line passes through the point (1, – 1/3, 1)
The position vector of this point is a =    – (1/3) +

The d.r. s of the line are i.e. 1/3, 1/6, – 1 i.e. 2, 1, -6 ≡ a, b, c
Hence, the direction vector of the line is b = 2   +  – 6

Now, the vector form of the equation of the line is given by

r = a + λ b

Where λ ∈ R, and is a scalar/parameter

The post Equation of Line in Space appeared first on The Fact Factor.

Example 01:

• Find the distance between the following pairs of points :
• (2, 3), (4, 1)

Let A(2, 3) ≡ (x₁, y₁) and B(4, 1) ≡ (x₂, y₂) be the points

By distance formula

AB² = (x₂ – x₁)² + (y₂ – y₁)²

∴  AB² = (4 – 2)² + (1 – 3)²  = (2)² + (-2)² = 4 + 4 = 8

∴  AB =√8  = 2√2 unit

Ans: The distance between given points is 2√2 unit

• (– 5, 7), (– 1, 3)

Let A(-5, 7) ≡ (x₁, y₁) and B(-1, 3) ≡ (x₂, y₂) be the points

By distance formula

AB² = (x₂ – x₁)² + (y₂ – y₁)²

∴  AB² = (-1 + 5)² + (3 – 7)²  = (4)² + (-4)² = 16 + 16 = 32

∴  AB =√32  = 4√2 unit

Ans: The distance between given points is 4√2 unit

• (6, 8), (– 9, – 12)

Let A(6, 8) ≡ (x₁, y₁) and B(-9, -12) ≡ (x₂, y₂) be the points

By distance formula

AB² = (x₂ – x₁)² + (y₂ – y₁)²

∴  AB² = (- 9 – 6)² + (-12 – 8)²  = (-15)² + (-20)² = 225 + 400 = 625

∴  AB =√625  = 25 unit

Ans: The distance between given points is 25 unit

• (-6, -1), (– 6, 11)

Let A(-6, -1) ≡ (x₁, y₁) and B(-6, 11) ≡ (x₂, y₂) be the points

By distance formula

AB² = (x₂ – x₁)² + (y₂ – y₁)²

∴  AB² = (- 6 + 6)² + (11 + 1)²  = (0)² + (12)² = 0 + 144 = 144

∴  AB =√144  = 12 unit

Ans: The distance between given points is 12 unit

• (a, b), (– a, – b)

Let A(a, b) ≡ (x₁, y₁) and B(-a, -b) ≡ (x₂, y₂) be the points

By distance formula

AB² = (x₂ – x₁)² + (y₂ – y₁)²

∴  AB² = (- a – a)² + (-b – b)²  = (- 2a)² + (- 2b)² = 4a² + 4b² = 4(a² + 4b²)

• (0, 0), (36, 15)

Let A(0, 0) ≡ (x₁, y₁) and B(36, 15) ≡ (x₂, y₂) be the points

By distance formula

AB² = (x₂ – x₁)² + (y₂ – y₁)²

∴  AB² = (36 – 0)² + (15 – 0)²  = (36)² + (15)² = 1296 + 225 = 1521

∴  AB =√1521  = 39 unit

Alternate Direct Method

One of the given point is origin O. Let A(36, 15) ≡ (x, y)

∴  OA² = x² + y² = (36)² + (15)²  = 1296 + 225 = 1521

∴  OA =√1521  = 39 unit

Ans: The distance between given points is 39 unit

Example 02:

• Using distance formula show that following sets of points are collinear
• A(0,4), B(2,10) and C(3,13)

Given points are A(0,4), B(2,10) and C(3,13)

Using distance formula

AB² = (2 – 0)² + (10 – 4)² = (2)² + (6)² = 4 + 36 = 40

∴  AB  = √40  =  2√10  unit …………….. (1)

BC² = (3 – 2)² + (13 – 10)² = (1)² + (3)² = 1 + 9 = 10

∴  BC  = √10   unit  …………….. (2)

AC² = (3 – 0)² + (13 – 4)² = (3)² + (9)² = 9 + 81 = 90

∴  AC  = √90  =  3√10  unit …………….. (3)

From equations (1), (2) and (3) we have

AC = AB + BC

∴ A – B – C

Hence points A, B and C are collinear

• P(5, 0), Q(10, -3) and R(-5, 6)

Given points are P(5, 0), Q(10, -3) and R(-5, 6)

Using distance formula

PQ² = (10 – 5)² + (-3 – 0)² = (5)² + (-3)² = 25 + 9 = 34

∴  PQ  = √34   unit   …………….. (1)

QR² = (-5 – 10)² + (6 + 3)² = (-15)² + (9)² = 225 + 81 = 306

∴  QR  = √306  =  3√34   unit …………….. (2)

PR² = (-5 – 5)² + (6 – 0)² = (-10)² + (6)² = 100+ 36 = 136

∴  PR  = √136  =  2√34   unit …………….. (3)

From equations (1), (2) and (3) we have

QR = PQ + PR

∴ Q – P – R

Hence points P, Q and R are collinear

• L(2, 5), M(5, 7) and N(8, 9)

Given points are L(2, 5), M(5, 7) and N(8, 9)

Using distance formula

LM² = (5 – 2)² + (7 – 5)² = (3)² + (2)² = 9 + 4 = 13

∴  LM  = √13   unit   …………….. (1)

MN² = (8 – 5)² + (9 – 7)² = (3)² + (2)² = 9 + 4 = 13

∴  MN  = √13  unit …………….. (2)

LN² = (8 – 2)² + (9 – 5)² = (6)² + (4)² = 36+ 16 = 52

∴  LN  = √52  =  2√13   unit …………….. (3)

From equations (1), (2) and (3) we have

LN = LM + MN

∴ L – M – N

Hence points L, M and N are collinear

• D(5, 1), E(1, -1) and F(11, 4)

Given points are D(5, 1), E(1, -1) and F(11, 4)

Using distance formula

DE² = (1 – 5)² + (-1 – 1)² = (-4)² + (-2)² = 16 + 4 = 20

∴  DE  = √20  =  2√5   unit   …………….. (1)

EF² = (11 – 1)² + (4 + 1)² = (10)² + (5)² = 100 + 25 = 125

∴  EF  = √125 =  5√5  unit …………….. (2)

DF² = (11 – 5)² + (4 – 1)² = (6)² + (3)² = 36+ 9 = 45

∴  DF  = √45  =  3√5  unit …………….. (3)

From equations (1), (2) and (3) we have

EF = DE + DF

∴ E – D – F

Hence points D, E and F are collinear

• A(1, 5), B(2, 3) and C(– 2, – 11)

Given points are A(1, 5), B(2, 3) and C(-2, -11)

Using distance formula

AB² = (2 – 1)² + (3 – 5)² = (1)² + (-2)² = 1 + 4 = 5

∴  AB  = √5   unit …………….. (1)

BC² = (-2 – 2)² + (-11 – 3)² = (-4)² + (-14)² = 16 + 196 = 212

∴  BC  = √212    unit  …………….. (2)

AC² = (-2 – 1)² + (-11 – 5)² = (-3)² + (-16)² = 9 + 256 = 265

∴  AC  = √265    unit …………….. (3)

From equations (1), (2) and (3) we have

AC ≠ AB + BC

Hence points A, B and C are not collinear

Example 03:

• Ashima, Bharti, and Camella are seated at A(3, 1), B(6, 4), and C(8, 6) respectively. Do you think they are seated in a line?

Given points are A(3, 1), B(6, 4) and C(8, 6)

Using distance formula

AB² = (6 – 3)² + (4 – 1)² = (3)² + (3)² = 9 + 9 = 18

∴  AB  = √18  = 3√2 unit …………….. (1)

BC² = (8 – 6)² + (6 – 4)² = (2)² + (2)² = 4 + 4 = 8

∴  BC  = √8  = 2√2   unit  …………….. (2)

AC² = (8 – 3)² + (6 – 1)² = (5)² + (5)² = 25 + 25 = 50

∴  AC  = √50   = 5√2  unit …………….. (3)

From equations (1), (2) and (3) we have

AC = AB + BC

∴ A – B – C

Hence points A, B and C are collinear.

Thus Ashima, Bharti and Camella are seated in a line

Example 04:

• The distance between the points (0, 0) and (x, 3) is 5. Find x.

Let A(0, 0) ≡ (x₁, y₁) and B(x, 3) ≡ (x₂, y₂) be the points.

Thus AB = 5

By distance formula

AB² = (x₂ – x₁)² + (y₂ – y₁)²

∴  5² = (x – 0)² + (3 – 0)²  = x² + 9

∴   25 = x² + 9

∴  x² = 16

∴  x = ± 4

Example 05:

• The distance between the points (a, 5) and (0, -3) is 4√5  unit. Find a.

Let A(a, 5) ≡ (x₁, y₁) and B(0, -3) ≡ (x₂, y₂) be the points.

Thus AB = 4√5

By distance formula

AB² = (x₂ – x₁)² + (y₂ – y₁)²

∴  (4√5)² = (0 – a)² + (-3 – 5)²  = (-a)² + (-8)² = a² + 64

∴  80 =  a² + 64

∴  a² = 16

∴  a = ± 4

Example 06:

• The distance between the points (8, -7) and (-4, a) is 13. Find a.

Let A(8, -7) ≡ (x₁, y₁) and B(-4, a) ≡ (x₂, y₂) be the points.

Thus AB = 5

By distance formula

AB² = (x₂ – x₁)² + (y₂ – y₁)²

∴  13² = (- 4 – 8)² + (a + 7)²  = (-12)² + a² + 14a + 49

∴  169 =  144 + a² + 14a + 49

∴  a² + 14a + 197 – 169 = 0

∴  a² + 14a + 24 = 0

∴  (a + 12)(a + 2) = 0

∴  a + 12 = 0 or a + 2 = 0

∴  a = -12 and a = -2

Example 07:

• Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Given P(2, -3) ≡ (x₁, y₁) and Q(10, y) ≡ (x₂, y₂) be the points.

Thus PQ = 10 units

By distance formula

PQ² = (x₂ – x₁)² + (y₂ – y₁)²

∴  10² = (10 – 2)² + (y + 3)²  = (8)² + y² + 6y + 9

∴  100 = 64 + y² + 6y + 9

∴  36 =  y² + 6y + 9

∴  y² + 6y + 9 – 36 = 0

∴  y² + 6y + – 27 = 0

∴  (y + 9)(y – 3) = 0

∴  y + 9 = 0 or y – 3 = 0

∴  y = -9 and y = 3

Example 08:

• Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Let P(2, –5) and Q(–2, 9) be the given points.

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

Given PA = QA

PA² = QA²

Using distance formula

(a – 2)² + (0 + 5)² = (a + 2)² + (0 – 9)²

∴  a² – 4a + 4 + 25  = a² + 4a + 4 + 81

∴   – 4a + 25  = + 4a + 81

∴   – 8a =  56

∴  a = – 7

Hence the required point is (-7, 0)

Example 09:

• Find a point on the y-axis which is equidistant from points A(6, 5) and B(– 4, 3).

A(6, 5) and B(– 4, 3) are given points.

Let the point on the y-axis be P(0, b)

Given PA = PB

PA² = PB²

Using distance formula

(0 – 6)² + (b – 5)² = (0 + 4)² + (b – 3)²

∴  36 + b² – 10 b + 25  = 16 + b² – 6b + 9

∴  36  – 10 b + 25  = 16  – 6b + 9

 ∴  – 4b =  25 – 61 = -36

∴  b = 9

Hence the required point is (0, 9)

Example 10:

• Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4)

Let P(3, 6) and Q(–3, 4) be the given points.

Given point A(x, y)

Given PA = QA

PA² = QA²

Using distance formula

(x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²

∴  x² – 6x + 9 + y² – 12y + 36  = x² + 6x + 9 + y² – 8y + 16

∴  x² – 6x + 9 + y² – 12y + 36  – x² – 6x – 9 – y² + 8y – 16 = 0

∴   – 12x – 4y + 20  = 0

∴  3x + y – 5 = 0

Hence the requiredrelation is 3x + y – 5 = 0

Example 11:

• Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

Let P(7, 1) and Q(3, 5) be the given points.

Given point A(x, y)

Given PA = QA

PA² = QA²

Using distance formula

(x – 7)² + (y – 1)² = (x – 3)² + (y – 5)²

∴  x² – 14x + 49 + y² – 2y + 1  = x² – 6x + 9 + y² – 10y + 25

∴  x² – 14x + 49 + y² – 2y + 1  – x² + 6x – 9 – y² + 10y – 25 = 0

∴   – 8x + 8y + 16  = 0

∴  x – y + 2 = 0

Hence the requiredrelation is x – y + 2 = 0

Example 12:

• If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also, find the distances QR and PR.

Given P(5, -3) and R(x, 6) be the given points.

Given point Q(0, 1)

Given PQ = RQ

PQ² = RQ²

Using distance formula

(5 – 0)² + (-3 – 1)² = (x – 0)² + (6 – 1)²

∴  (5)² + (-4)² = (x)² + (5)²

∴  25 + 16 = x² + 25

∴  x²  = 16

∴  x = ± 4

When R(4, -3)

QR² = (4 – 0)² + (-3 – 1)² = (4)² + (-4)² = 16 + 16 = 32

∴  QR  = √32  =  4√2   unit   …………….. (1)

PR² = (4 – 5)² + (-3 + 3)² = (-1)² + (0)² = 1 + 0 = 1

∴  PR  = √1  =  1   unit   …………….. (1)

When R(-4, -3)

QR² = (-4 – 0)² + (-3 – 1)² = (-4)² + (-4)² = 16 + 16 = 32

∴  QR  = √32  =  4√2   unit   …………….. (1)

PR² = (-4 – 5)² + (-3 + 3)² = (-9)² + (0)² = 81 + 0 = 81

∴  PR  = √81  =  9   unit   …………….. (1)

Example 13:

• If P (6, -1), Q(1, 3), and R(x, 8) are the points and PQ = QR,  find the values of x.

Given P(6, -1) and R(x, 8) be the given points.

Given point Q(1, 3)

Given PQ = RQ

PQ² = RQ²

Using distance formula

(1 – 6)² + (3 + 1)² = (1 – x)² + (3 – 8)²

∴  (-5)² + (-4)² = x² – 2x + 1 + (-5)²

∴  25 + 16 = x² – 2x + 1 +25

∴  x²  – 2x + 1 – 16 = 0

∴  x²  – 2x + – 15 = 0

∴  (x + 3)(x – 5) = 0

∴ x + 3 = 0 and x – 5 = 0

∴ x = – 3 and x = 5

The post Distance Formula appeared first on The Fact Factor.

In this article, we shall discuss problems to find the condition when the relation between slopes of lines represented by a joint equation is given.

Example – 18:

• If the slope of one of the lines given by ax² + 2hxy + by² = 0 is k times the slope of the other, prove that 4kh² = ab(1+ k)².
• Solution:

The given joint equation is ax² + 2hxy + by² = 0

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b, and  m₁ . m₂ = a/b

Given the relation between slopes of lines

m₁ = km₂

Now, m₁ + m₂ = – 2h/b

∴  km₂ + m₂ =- 2h/b

∴ (k + 1)m₂ =- 2h/b

∴  m₂ = – 2h/b(k + 1)

Now, m₁ . m₂ = a/b

∴  km₂ . m₂ = a/b

∴  k(m₂)² = a/b

∴  (m₂)² = a/bk

4kh² = ab(1+ k)² (proved as required)

Example – 19:

• If the slope of one of the lines given by ax² + 2hxy + by² = 0 is 5 times the slope of the other, prove that 5h² = 9ab.
• Solution:

The given joint equation is ax² + 2hxy + by² = 0

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b, and  m₁ . m₂ = a/b

Given the relation between slopes of lines

m₁ = 5m₂

Now, m₁ + m₂ = – 2h/b

∴  5m₂ + m₂ =- 2h/b

∴    6m₂ =- 2h/b

∴  m₂ = – h/3b

Now, m₁ . m₂ = a/b

∴  5m₂ . m₂ = a/b

∴  5(m₂)² = a/b

5h² = 9ab  (proved as required)

Example – 20:

• If the slope of one of the lines given by ax² + 2hxy + by² = 0 is 4 times the slope of the other, prove that 16h² = 25ab.
• Solution:

The given joint equation is ax² + 2hxy + by² = 0

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b, and  m₁ . m₂ = a/b

Given the relation between slopes of lines

m₁ = 4m₂

Now, m₁ + m₂ = – 2h/b

∴  4m₂ + m₂ =- 2h/b

∴    5m₂ =- 2h/b

∴  m₂ = – 2h/5b

Now, m₁ . m₂ = a/b

∴  4m₂ . m₂ = a/b

∴  4(m₂)² = a/b

16h² = 25ab  (proved as required)

Example – 21:

• If the slope of one of the lines given by ax² + 2hxy + by² = 0 is 3 times the slope of the other, prove that 3h² = 4ab.
• Solution:

The given joint equation is ax² + 2hxy + by² = 0

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b, and  m₁ . m₂ = a/b

Given the relation between slopes of lines

m₁ = 3m₂

Now, m₁ + m₂ = – 2h/b

∴  3m₂ + m₂ =- 2h/b

∴    4m₂ =- 2h/b

∴  m₂ = – h/2b

Now, m₁ . m₂ = a/b

∴  3m₂ . m₂ = a/b

∴  3(m₂)² = a/b

3h² = 4ab  (proved as required)

Example – 22:

• Find the condition that slope of one of the lines represented by ax² + 2hxy  + by² = 0 is square of the slope of another line.
• Solution:

The given joint equation is ax² + 2hxy + by² = 0

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b, and  m₁ . m₂ = a/b

Given the relation between slopes of lines

m₁ = (m₂)²

Now, m₁ . m₂ = a/b

∴  (m₂)² . m₂ = a/b

∴  (m₂)³ = a/b

Now, m₁ + m₂ = – 2h/b

∴  (m₂)² + m₂ = – 2h/b

Making cube of both sides

((m₂)² + m₂)³ = (- 2h/b)³

Multiplying both sides by b³

a²b – 6abh + ab² = -8h³

a²b  + ab² + 8h³ = 6abh

This is the required condition

Example – 23:

• Show that one of the straight line  given by ax² + 2hxy + by² = 0 bisects an angle between the co-ordinate axes if and only if  (a + b)² = 4h².
• Solution:

The given joint equation is ax² + 2hxy + by² = 0

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b, and  m₁ . m₂ = a/b

Now given that One of the straight line represented by the joint equation bisects the angle between the co-ordinate axes. Thus the slope of the line is ± 1. let m₂ = ± 1

Example – 24:

• If the straight lines represented by a joint equation  ax² + 2hxy + by² = 0 make angles of equal measures with the co-ordinate axes. Then prove that a = ± b.
• Solution:

Let α be the angle made by the lines with coordinate axes.

Let OA be the line making an angle α with the x-axis and

OB be the line making an angle α with y-axis.

The angle made by the line with the positive direction of x-axis is α. Hence its slope is

m₁ = tan α

The angle made by OB with the positive direction of x-axis is (π/2 ± α). Hence its slope is

m₂ = tan (π/2 ± α) = ± cot α

The given joint equation is ax² + 2hxy + by² = 0

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b, and  m₁ . m₂ = a/b

Science > Mathematics > Pair of Straight Lines > To Find the Condition When Relation Between Slopes is Given

The post To Find Condition When Relation Between Slopes f lines is Given appeared first on The Fact Factor.

In this article, we shall study to solve problems to find the value of constant or to prove the given relation when the relation between slopes of the line represented by the joint equation is given.

Remember:

• If m₁ and m₂ are the slopes of the two lines represented by joint equation  ax² + 2hxy  + by² = 0. Then m₁. m₂ = a/b ; m₁ + m₂ = -2h/b  and remember (m₁ – m₂) 2 =  (m₁+ m₂)² – 4m₁.m₂.

Algorithm:

1. Write given of joint equation.
2. Compare with the standard equation.
3. Find values of a, h and b.
4. Use above relations between slopes
5. Solve equations and get the value of constant

Example – 01:

• Show that the equation 3x² – 4xy +  y² = 0 represents the pair of lines whose slopes differ by 2.
• Solution:

The given joint equation is 3x² – 4xy + y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = 3, 2h = – 4, and b = 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = -(-4)/1 = 4 and  m₁ . m₂ = a/b = 3/1 = 3

Using  (m₁ – m₂)² =  (m₁ + m₂)² –  4 m₁ . m₂

∴   (m₁ – m₂)² =  (4)² –  4 (3)

∴   (m₁ – m₂)² =  16 – 12

∴   (m₁ – m₂)² =  4

Taking square roots of both sides we get,

| m₁ – m₂| = 2

Hence the slopes differ by 2.

Example – 02:

• Show that the equation 77x² – 36xy +  4y² = 0 represents the pair of lines whose slopes differ by 2.
• Solution:

The given joint equation is 77x² – 36xy +  4y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = 77, 2h = – 36, and b = 4

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = -(-36)/4 = 9 and  m₁ . m₂ = a/b = 77/4

Using  (m₁ – m₂)² =  (m₁ + m₂)² –  4 m₁ . m₂

∴   (m₁ – m₂)² =  (9)² –  4 (77/4)

∴   (m₁ – m₂)² =  81 – 77

∴   (m₁ – m₂)² =  4

Taking square roots of both sides we get,

| m₁ – m₂| = 2

Hence the slopes differ by 2.

Example – 03:

• Show that the difference of the slopes of the two lines represented by equation x²(tan²θ + cos²θ)  – 2xy tanθ + y²sin²θ = 0 is 2.
• Solution: 

The given joint equation is x²(tan²θ + cos²θ)  – 2xy tanθ + y²sin²θ = 0

Comparing with ax² + 2hxy + by² = 0

Here a = (tan²θ + cos²θ), 2h = – 2tanθ, and b = sin²θ

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = -(- 2tanθ)/sin²θ = 2tanθ)/sin²θ

and  m₁ . m₂ = (tan²θ + cos²θ)/sin²θ

Using  (m₁ – m₂)² =  (m₁ + m₂)² –  4 m₁ . m₂

Hence the slopes differ by 2.

Example – 04:

• Show that the difference of the slopes of the two lines represented by equation (tan²θ + 3)(tan²θ – 1)x² – 2xy sec²θ   +  y² = 0 is 4.
• Solution:

The given joint equation is (tan²θ + 3)(tan²θ – 1)x² – 2xy sec²θ   +  y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = (tan²θ + 3)(tan²θ – 1), 2h = – 2sec²θ, and b = 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = -(- 2sec²θ)/1 =  2sec²θ

and  m₁ . m₂ = (tan²θ + 3)(tan²θ – 1)/1 = (tan²θ + 3)(tan²θ – 1)

Using  (m₁ – m₂)² =  (m₁ + m₂)² –  4 m₁ . m₂

Hence the slopes differ by 4.

Example – 05:

• Find λ, if the difference of the slopes of the lines λx² + 6xy –  4y² = 0 is equal to their product.
• Solution:

The given joint equation is λx² + 6xy –  4y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = λ, 2h = 6, and b = – 4

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = -6/-4 = 3/2 and  m₁ . m₂ = a/b = λ/-4  = – λ/4

Given the relation between slopes

| m₁ – m₂| = m₁ . m₂

squaring both sides

 (m₁ – m₂)² =  (m₁ . m₂)²

 ∴   (m₁ + m₂)² –  4 m₁ . m₂=  (m₁ . m₂)²

∴   (3/2)² –  4 (- λ/4)=  (- λ/4)²

∴   9/4 +  λ =  λ²/16

Multiplying both sides of equation by 16

∴   36 +  16λ =  λ²

∴  λ² –  16λ  – 36 =  0

∴  (λ – 18)(λ + 2) =  0

∴   λ = 18 or λ = – 2

Example – 06:

• Find k, if the difference of the slopes of the lines  3x² + kxy  – y² = 0 is 4.
• Solution:

The given joint equation is 3x² + kxy  – y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = 3, 2h = k, and b = – 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = -k/-1 = k and  m₁ . m₂ = a/b  = 3/-1 = – 3

Given the relation between slopes

| m₁ – m₂| = 4

squaring both sides

 (m₁ – m₂)² = 16

 ∴   (m₁ + m₂)² –  4 m₁ . m₂=  16

∴   (k)² –  4 (- 3)=  16

∴   k² + 12 =  16

∴   k²  = 4

∴   k = ± 2

Example – 07:

• Find k if the slopes of lines given by kx² + 5xy  + y² = 0 differ by 1.
• Solution:

The given joint equation is kx² + 5xy  + y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = k, 2h = 5, and b = 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – 5/1 = – 5 and  m₁ . m₂ = a/b = k/1 = k

Given the relation between slopes

| m₁ – m₂| = 1

squaring both sides

 (m₁ – m₂)² = 1

 ∴   (m₁ + m₂)² –  4 m₁ . m₂= 1

 ∴   (-5)² –  4 k=  1

∴  25 –  4 k=  1

∴   –  4 k=  – 24

∴   k = 6

Example – 08:

• Find k, if the slope of one of the lines given by kx² + 4xy  – y² = 0 exceeds the slope of other by 8.
• Solution:

The given joint equation is kx² + 4xy  – y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = k, 2h = 4, and b = – 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – 4/-1 =  4 and  m₁ . m₂ = a/b  = k/-1 = – k

Given the relation between slopes

| m₁ – m₂| = 8

squaring both sides

 (m₁ – m₂)² = 64

 ∴   (m₁ + m₂)² –  4 m₁ . m₂= 64

 ∴   (4)² –  4(- k)=  64

∴  16 +  4 k=  64

∴   – 4 k= 48

∴   k = 12

Example – 09:

• Find k, if the difference between the slopes of the lines  12x² + k xy  – y² = 0 is 7.
• Solution:

The given joint equation is 12x² + k xy  – y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = 12, 2h = k, and b = – 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – k/-1 = k and  m₁ . m₂ = a/b = 12/-1 = – 12

Given the relation between slopes

| m₁ – m₂| = 7

squaring both sides

 (m₁ – m₂)² = 49

 ∴   (m₁ + m₂)² –  4 m₁ . m₂= 49

 ∴   (k)² –  4(- 12)=  49

 ∴   k² + 48 =  49

∴  k² = 1

∴   k = ± 1

Example – 10:

• Find k, if the slope of one of the lines given by 3x² + 4xy  + ky² = 0 is three times the slope of the other.
• Solution:

The given joint equation is 3x² + 4xy  + ky² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = 3, 2h = 4, and b = k

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – 4/k and  m₁ . m₂ = a/b = 3/k

Given the relation between slopes

m₁ = 3m₂

∴  k² = k

∴  k² – k = 0

∴  k (k – 1) = 0

∴  k = 0 and k = 1

But k cannot be zero.

∴ k = 1

Example – 11:

• Find k, if the slope of one of the lines given by kx² + 4xy  – y² = 0 is three times the slope of the other.
• Solution:

The given joint equation is kx² + 4xy  – y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = k, 2h = 4, and b = – 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – 4/-1 = 4, and  m₁ . m₂ = a/b = k/-1 = – k

Given the relation between slopes

m₁ = 3m₂

Now, m₁ + m₂ = 4

∴  3m₂ + m₂ = 4

∴  4m₂ = 4

∴  m₂ = 1

∴  m₁ = 3m₂ = 3 x 1 = 3

Now, m₁ . m₂ =  – k

∴  (3)(1) = – k

∴  k = – 3

Example – 12:

• Find k, if the slope of one of the lines given by 4x² + kxy  + y² = 0 is four times the slope of the other.
• Solution:

The given joint equation is 4x² + kxy  + y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = 4, 2h = k, and b = 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – k/1 = – k, and  m₁ . m₂ = a/b = 4/1 = 4

Given the relation between slopes

m₁ = 4m₂

Now, m₁ + m₂ = – k

∴  4m₂ + m₂ = – k

∴  5m₂ = – k

∴  m₂ = – k/5

Now, m₁ . m₂ = 4

∴  4m₂ . m₂ = 4

∴  (m₂)² = 1

∴  (- k/5)² = 1

∴  (- k/5) = ± 1

∴  k = ± 5

Example – 13:

• Find k, if the slope of one of the lines given by 5x² + kxy  + y² = 0 is five times the slope of the other.
• Solution:

The given joint equation is 5x² + kxy  + y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = 5, 2h = k, and b = 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – k/1 = – k, and  m₁ . m₂ = a/b = 5/1 = 5

Given the relation between slopes

m₁ = 5m₂

Now, m₁ + m₂ = – k

∴  5m₂ + m₂ = – k

∴  6m₂ = – k

∴  m₂ = – k/6

Now, m₁ . m₂ = 5

∴  5m₂ . m₂ = 5

∴  (m₂)² = 1

∴  (- k/6)² = 1

∴  (- k/6) = ± 1

∴  k = ± 6

Example – 14:

• Find k if the sum of the slopes of lines given by kx² + 8xy + 5 y² = 0 is twice the product of the slopes.
• Solution:

The given joint equation is kx² + 8xy + 5 y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = k, 2h = 8, and b = 5

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – 8/5, and  m₁ . m₂ = a/b = k/5

Given the relation between slopes

m₁ + m₂ = 2 m₁ . m₂

-8/5 = 2 x k/5

∴  k = – 4

Example – 15:

• Find k if the sum of the slopes of lines given by 2x² + kxy -3y² = 0 is equal to the product of the slopes.
• Solution:

The given joint equation is 2x² + kxy -3y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = 2, 2h = k, and b = -3

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – k/-3 = k/3, and  m₁ . m₂ = a/b = 2/-3 = – 2/3

Given the relation between slopes

m₁ + m₂ = m₁ . m₂

k/3 = (-2/3)

∴  k = – 2

Example – 16:

• Find k, if the sum of the slopes of lines given by 3x² + kxy  + y² = 0 is zero.
• Solution:

The given joint equation is 2x² + kxy -3y² = 0

Comparing with ax² + 2hxy + by² = 0

Here a = 3, 2h = k, and b = 1

Let m₁ and m₂ be the slopes of the two lines represented by given joint equation

 (m₁ + m₂) = – 2h/b = – k/1 = – k, and  m₁ . m₂ = a/b = 3/1 = 3

Given the relation between slopes

m₁ + m₂ = 0

∴  – k = 0

∴  k = 0

Example – 17:

• Find k, if the slope of one of the lines given by 3x² – 4xy  + ky² = 0 is 1.
• Solution:

Given joint equation of lines is 3x² – 4xy  + ky² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 3, 2h = – 4 and b= k

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   km² – 4m + 3 = 0   ……… (1)

Slope of one of the line is 1

Substituting m = 1 in equation (1) we get

k(1)² – 4(1) + 3 = 0

∴   k – 1 = 0

∴   k = 1

Science > Mathematics > Pair of Straight Lines > To Find Value of Constant When Relation Between Slopes is Given

The post To Find Value of Constant When Relation Between Slopes is Given appeared first on The Fact Factor.

Standard Equation of Circle:

By distance formula

Let P(x, y) be any point on a circle having centre at O(0, 0) and radius ‘a’.
Then, OP = radius of a circle = a
By distance formula
As this contains less number of constants, the equation is called the standard equation of a circle.

Centre Radius Form:

By distance formula

Let P(x, y) be any point on a circle having centre at C(h, k) and radius ‘r’.
Then, CP = radius of a circle = r
By distance formula
This form of the equation of a circle is called centre radius form.

Equation of Circle When Centre and Radius are Given

Algorithm:

1. Write given centre ≡ (h, k) and given radius = r
2. Use centre radius form (x – h)² + (y – k)² = r²
3. Simplify the equation and write it in standard form ax² + by² + 2gx + 2fy + c = 0

Example – 01:

• Find the equation of circle having centre at (2, -3) and radius 5.
• Solution:

Given centre (2, -3) ≡ (h, k) and radius = r = 5

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x – 2)² + (y + 3)² = 5²

∴  x² – 4x + 4 + y² + 6y + 9 = 25

∴  x² + y² – 4x + 6y + 13 – 25 = 0

∴  x² + y² – 4x + 6y – 12 = 0

Ans: The required equation of circle is x² + y² – 4x + 6y – 12 = 0

Example – 02:

• Find equation of circle having centre at origin and radius 4.
• Solution:

Given centre (0, 0) ≡ (h, k) and radius = r = 4

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x – 0)² + (y + 0)² = 4²

∴  x²  + y² = 16

Ans: The required equation of circle is x² + y²  = 16

Example – 03:

• Find the equation of circle having centre at (3, -2) and radius 5.
• Solution:

Given centre (3, -2) ≡ (h, k) and radius = r = 5

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x – 3)² + (y + 2)² = 5²

∴  x² – 6x + 9 + y² + 4y + 4 = 25

∴  x² + y² – 6x + 4y + 13 – 25 = 0

∴  x² + y² – 6x + 4y – 12 = 0

Ans: The required equation of circle is x² + y² – 6x + 4y – 12 = 0

Example – 04:

• Find the equation of circle having centre at (-3, -2) and radius 6.
• Solution:

Given centre (-3, -2) ≡ (h, k) and radius = r = 6

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x + 3)² + (y + 2)² = 6²

∴  x² + 6x + 9 + y² + 4y + 4 = 36

∴  x² + y² + 6x + 4y + 13 – 36 = 0

∴  x² + y² + 6x + 4y – 23 = 0

Ans: The required equation of circle is x² + y² + 6x + 4y  – 23 = 0

Example – 05:

• Find the equation of circle having centre at (a cosα, a sinα) and radius a.
• Solution:

Given centre (a cosα, a sinα) ≡ (h, k) and radius = r = 6

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

(x – a cosα)² + (y – a sinα)² = a²

∴  x² – 2xa cosα + a² cos²α + y² – 2ya sinα + a² sin²α = a²

∴  x² + y²  – 2ax cosα  – 2ay sinα + a² cos²α  + a² sin²α = a²

∴  x² + y²  – 2ax cosα  – 2ay sinα + a² (cos²α  + sin²α) – a²  = 0

∴  x² + y²  – 2ax cosα  – 2ay sinα + a² (1) – a²  = 0

∴  x² + y²  – 2acosα.x  – 2a sinα.y = 0

Ans: The required equation of circle is x² + y²  – 2acosα.x  – 2a sinα.y = 0

Example – 06:

• Find the equation of circle having centre at (a, b) and radius  √a²+b²
• Solution:

Given centre (a, b) ≡ (h, k) and radius = r = √a²+b² 

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x – a)² + (y – b)² = (√a²+b²  )²

∴  x² – 2ax + a² + y² – 2by + b² = a² + b²

∴  x² + y² – 2ax – 2by + a²  + b² – a² – b² = 0

∴  x² + y² – 2ax – 2by  = 0

Ans: The required equation of circle is x² + y² – 2ax – 2by  = 0

Equation of Circle When Centre and Point on the Circle are Given:

Algorithm:

1. Write given centre C ≡ (h, k) and given point say P (x, y)
2. Use distance formula to find CP represent it as radius = r
3. Use centre radius form (x – h)² + (y – k)² = r²
4. Simplify the equation and write it in standard form ax² + by² + 2gx + 2fy + c = 0

Example 07:

• Find the equation of circle having centre at (-3, 1) and passing through (5, 2)
• Solution:

Given centre C(-3, 1) ≡ (h, k) and  point on circle P(5, 2)

By distance formula

CP² = (5 + 3)² + (2 – 1)² = 64 + 1 = 65

∴  CP = √65

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x + 3)² + (y – 1)² = (√65)²

∴  x² + 6x + 9 + y² – 2y + 1 = 65

∴  x² + y² + 6x – 2y + 10 – 65 = 0

∴  x² + y² + 6x – 2y – 55 = 0

Ans: The required equation of circle is x² + y² + 6x – 2y – 55 = 0

Example 08:

• Find the equation of circle having centre at (2, -1) and passing through (3, 6)
• Solution:

Given centre C(2, -1) ≡ (h, k) and  point on circle P(3, 6)

By distance formula

CP² = (3 – 2)² + (6 + 1)² =1 + 49 = 50

∴  CP = √50

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x – 2)² + (y + 1)² = (√50)²

∴  x² – 4x + 4 + y² + 2y + 1 = 50

∴  x² + y² – 4x + 2y + 5 – 50 = 0

∴  x² + y² – 4x + 2y – 45 = 0

Ans: The required equation of circle is x² + y² – 4x + 2y – 45 = 0

Example 09:

• Find the equation of circle having centre at (2, -3) and passing through P(-3, 5)
• Solution:

Given centre C(2, – 3) ≡ (h, k) and  point on circle (-3, 5)

By distance formula

CP² = (-3 – 2)² + (5 + 3)² = 25 + 68 = 89

∴  CP = √89

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x – 2)² + (y + 3)² = (√89)²

∴  x² – 4x + 4 + y² + 6y + 9 = 89

∴  x² + y² – 4x + 6y + 13 – 89 = 0

∴  x² + y² – 4x + 6y – 76 = 0

Ans: The required equation of circle is x² + y² – 4x + 6y – 76 = 0

Example 10:

• Find equation of circle passing through (-3, 2) and having centre at the point of intersection of lines x – 2y = 4 and 2x + 5y + 1 = 0
• Solution:

Solving the equations x – 2y = 4 and 2x + 5y + 1 = 0 we get x = 2 and y = -1

Thus centre C(2, – 1) ≡ (h, k) and  point on circle P(-3, 2)

By distance formula

CP² = (-3 – 2)² + (2 + 1)² = 25 + 9 = 34

∴  CP = √34

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x – 2)² + (y + 1)² = (√34)²

∴  x² – 4x + 4 + y² + 2y + 1 = 34

∴  x² + y² – 4x + 2y + 5 – 34 = 0

∴  x² + y² – 4x + 2y – 29 = 0

Ans: The required equation of circle is x² + y² – 4x + 2y – 29 = 0

Example 11:

• Find equation of circle passing through intersection of x + 3y = 0 and 2x – 7y = 0 and having centre at the point of intersection of lines x + y + 1 = 0 and x – 2y + 4 = 0
• Solution:

Solving the equations x + 3y = 0 and 2x – 7y = 0 we get x = 0 and y = 0

Hence the circle passes through O(0, 0)

Solving the equations x + y + 1 = 0 and x – 2y + 4 = 0 we get x = -2 and y = 1

Hence the centre of circle is C(-2, 1)

Thus centre C(-2, 1) ≡ (h, k) and  point on circle O(0, 0)

By distance formula

CO² = (0 + 2)² + (0 – 1)² = 4 + 1 = 5

∴  CO = √5

By centre radius form, the equation of circle is given by

(x – h)² + (y – k)² = r²

∴  (x + 2)² + (y – 1)² = (√5)²

∴  x² + 4x + 4 + y² – 2y + 1 = 5

∴  x² + y² + 4x – 2y + 5 – 5 = 0

∴  x² + y² + 4x – 2y = 0

Ans: The required equation of circle is x² + y² + 4x – 2y = 0

Find equation of a circle having centre at (- 4, 0) and radius 5. Ans: x2 + y2 + 8x – 34 = 0
6. Find equation of a circle having centre at (7, 2) and radius 3. Ans: x2 + y2 – 14x – 4y + 44 = 0
7. Find equation of a circle having centre at (- 5, 1) and radius 2. Ans: x2 + y2 + 10x – 2y + 14 = 0
8. Find equation of a circle having centre at (1, -1) and passing through (3, 2) (M – 78)
Ans: x2 + y2 – 2x + 2y – 11 = 0
9. Find equation of a circle having centre at (2, -1) and passing through (5, 3) (O – 81)
Ans: x2 + y2 – 4x + 2y – 20 = 0.
10. Find equation of a circle having centre at (2, -3) and passing through (-1, 2) (M – 98)
Ans: x2 + y2 – 4x + 6y – 21 = 0
11. Find equation of a circle having centre at (1, -2) and passing through intersection of lines 3x + y = 14 and 2x + 5y = 18. Ans: x2 + y2 – 2x + 4y – 20 = 0
12. Find equation of a circle having centre at (- 4, 3) and Y – axis is tangent to it. (O – 78)
Ans: x2 + y2 + 8x – 6y + 9 = 0
13. Find equation of a circle having centre at (2, – 6) and touching X – axis.
Ans: x2 + y2 – 4x+ 12y + 4 = 0
14. Find equation of a circle having centre at (2, 3) and touching X – axis.
Ans: x2 + y2 – 4x – 6y + 4 = 0
15. Find equation of a circle having centre at (7, – 2) and touching X – axis.
Ans: x2 + y2 – 14x + 4y + 49 = 0
16. Find equation of a circle which touches X – axis at (-1, 0) and have radius 5.
Ans:

The post Equation of Circle (Centre Radius Form) appeared first on The Fact Factor.

In this article, we shall study problems to find the condition when one of the lines represented by a homogeneous equation is given.

ALGORITHM :

• Write the auxiliary equation of joint equation.
• Find the slope of given line m.
• Substitute value of m in auxillary equation.
• Simplify and get the condition.

Example – 01:

• If the joint equation of lines is ax² + 2hxy  + by² = 0. Find the condition that the line 5x + 3y = 0 is one of them.
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

One of the line is 5x + 3y = 0. Its slope is – 5/3

Now – 5/3 must be one of the roots of the auxiliary equation (1),

Substituting m = – 5/3 in equation (1)

Multiplying both sides of the equation by 9

25 b -15 h + 9a = 0

9a + 25 b -15 h  = 0

This is the required condition.

Example – 02:

• If the joint equation of lines is  ax² + 2hxy  + by² = 0. Find the condition that the line 3x – y = 0 is one of them.
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

One of the line is 3x – y = 0. Its slope is – 3/-1 = 3

Now 3 must be one of the roots of the auxiliary equation (1),

b(3)² + 2h(3) + a = 0

∴   9b + 6h + a = 0

∴   a + 9b + 6h = 0

This is the required condition.

Example – 03:

• If the joint equation of lines is ax² + 2hxy  + by² = 0. the line 4x – 3y = 0 coincides with one of them. Show that   16 b + 24 h + 9a = 0.
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

One of the line is 4x – 3y = 0. Its slope is – 4/-3 = 4/3

Now 3 must be one of the roots of the auxiliary equation (1),

b(4/3)² + 2h(4/3) + a = 0

∴  b(16/9) + 2h(4/3) + a = 0

Multiplying both sides of the equation by 9

16b + 24 h + 9a = 0, Proved as required.

Example – 04:

• Find the condition that one of the line represented by ax² + 2hxy  + by² = 0 is y = mx
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

One of the line is y = mx . Its slope is m

Now m must be one of the roots of the auxiliary equation (1),

b(m)² + 2h(m) + a = 0

∴  bm² + 2hm + a = 0

This is the required condition..

Example – 05:

• Find the condition that one of the line represented by ax² + 2hxy  + by² = 0 is px + qy = 0.
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

One of the line is px + qy = 0. Its slope is – p/q

Now – p/q must be one of the roots of the auxiliary equation (1),

This is the required condition

Example – 06:

• Find the condition that one of the line represented by ax² 2hxy  + by² = 0. is 3x – 2y = 0
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

One of the line is 3x – 2y = 0. Its slope is – 3/-2 = 3/2

Now 3/2 must be one of the roots of the auxiliary equation (1),

b(3/2)² + 2h(3/2) + a = 0

∴  b(9/4) + 2h(3/2) + a = 0

Multiplying both sides of the equation by 4

9b + 12 h + 4a = 0

This is the required condition

 Example – 07:

• Find the condition that the line 4x + 5y = 0 coincides with one of the line represented by ax² + 2hxy  + by² = 0. OR  If the line 4x + 5y = 0 coincides with one of the line epresented by ax² + 2hxy  + by² = 0, prove that 25a – 40h + 16 b = 0.
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

One of the line is 4x + 5y = 0. Its slope is – 4/5

Now – 4/5 must be one of the roots of the auxiliary equation (1),

b(-4/5)² + 2h(-4/5) + a = 0

∴  b(16/25) + 2h(-4/5) + a = 0

Multiplying both sides of the equation by 25

16b – 40 h + 25a = 0

25a – 40h + 16 b = 0. Proved as required

Example – 08:

• If the joint equation of lines is ax² + 2hxy  + by² = 0. Find the condition that the line 3x + 4y = 7 is perpendicular to one of them.
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

Given line is 3x + 4y = 7. Its slope is – 3/4

As one of the lines is perpendicular to given line

Hence slope of one of the line represented by joint equation is 4/3

Now 4/3 must be one of the roots of the auxiliary equation (1),

b(4/3)² + 2h(4/3) + a = 0

∴  b(16/9) + 2h(4/3) + a = 0

Multiplying both sides of the equation by 9

16b + 24h + 9a = 0

9a + 16b + 24h = 0

This is the required condition

Example – 09:

• If the joint equation of lines is ax² + 2hxy  + by² = 0. Find the condition that the line x + 2y = 0 is perpendicular to one of them.
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

Given line is x + 2y = 0. Its slope is – 1/2

As one of the lines is perpendicular to given line

Hence slope of one of the line represented by joint equation is 2

Now 2 must be one of the roots of the auxiliary equation (1),

b(2)² + 2h(2) + a = 0

4b + 4h + a = 0

a + 4b + 4h = 0

This is the required condition

Example – 10:

• If the joint equation of lines is ax² + 2hxy  + by² = 0. Find the condition that the line 3x + y = 0 is perpendicular to one of them.
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

Given line is 3x + y = 0. Its slope is – 3/1 = -3

As one of the lines is perpendicular to given line

Hence slope of one of the line represented by joint equation is 1/3

Now 1/3 must be one of the roots of the auxiliary equation (1),

b(1/3)² + 2h(1/3) + a = 0

b(1/9) + 2h(1/3) + a = 0

Multiplying both sides by 9

b + 6h + 9a = 0

9a + b + 6h = 0

This is the required condition

Example – 11:

• If the joint equation of lines is ax² + 2hxy  + by² = 0. Find the condition that the line px + qy = 0 is perpendicular to one of them.
• Solution:

Given joint equation of lines is ax² + 2hxy  + by² = 0.

Its auxiliary equation is

bm² + 2hm + a = 0  ……… (1)

Given line is px + qy = 0. Its slope is – p/q

As one of the lines is perpendicular to given line

Hence slope of one of the line represented by joint equation is q/p

Now q/p must be one of the roots of the auxiliary equation (1),

b(q/p)² + 2h(q/p) + a = 0

b(q²/p²) + 2h(q/p) + a = 0

Multiplying both sides by p²

bq² + 2pqh + ap² = 0

ap² + bq² + 2pqh = 0

This is the required condition

Science > Mathematics > Pair of Straight Lines > Condition That Line is Represented by Homogeneous Equation

The post Condition That Line is Represented by Homogeneous Equation appeared first on The Fact Factor.

In this article, we shall study the use of the auxiliary equation of par of lines to find the required condition.

Algorithm:

• Write the auxiliary equation of the joint equation.
• Find the slope of given line m.
• Substitute value of m in auxiliary equation.
• Simplify and get the value of constant.

Example 01:

• Find the value of ‘k’ if one of the line given by 6x² + kxy + y² = 0 is 2x + y = 0.
• Solution:

Given joint equation of lines is 6x² + kxy + y² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 6, 2h = k and b= 1

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (1)m² + km + 6 = 0

∴   m² + km + 6 = 0 ……… (1)

One of the line is 2x + y = 0. Its slope is – 2/1 = -2

Now – 2 must be one of the roots of the auxiliary equation (1),

Substituting m = – 2 in equation (1)

(-2)² + k(-2) + 6 = 0

∴   4 – 2k + 6 = 0

 ∴   – 2k  = – 10

∴   k = 5

Example 02:

• Find λ, if  2x + 5y = 0 coincides with one of the lines x² – λxy +  5y² = 0.
• Solution:

Given joint equation of lines is x² – λxy +  5y² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 1, 2h = – λ and b= 5

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (5)m² – λm + 1 = 0

∴   5m² – λm + 1 = 0 ……… (1)

One of the line is 2x + 5y = 0. Its slope is – 2/5

Now – 2/5 must be one of the roots of the auxiliary equation (1),

Substituting m = – 2/5 in equation (1)

5(-2/5)² –  λ(-2/5) + 1 = 0

∴  5(4/25) –  λ(-2/5) + 1 = 0

 ∴  (4/5) –  λ(-2/5) + 1 = 0

Multiplying both sides of equation by 5

∴  4 +  2λ  + 5 = 0

∴   2λ = – 9

∴  λ = – 9/2

Example 03:

• Find k, if the one of the lines given by 2x² – xy  + ky² = 0 is x – 3y = 0 .
• Solution:

Given joint equation of lines is 2x² – xy  + ky² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 2, 2h = – 1 and b= k

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (k)m² – 1m + 2 = 0

∴   km² –  m + 2 = 0 ……… (1)

One of the line is x – 3y = 0. Its slope is – 1/-3 = 1/3

Now 1/3 must be one of the roots of the auxiliary equation (1),

Substituting m = 1/3 in equation (1)

k(1/3)² –  (1/3) + 2 = 0

∴  K(1/9) –  1/3 + 2 = 0

Multiplying both sides of equation by 9

∴  k – 3 + 18 = 0

∴   k = -15

Example 04:

• Find k, if the one of the lines given by kx² + 3xy  – y² = 0 is 2x + y = 0
• Solution:

Given joint equation of lines is kx² + 3xy  – y² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = k, 2h = 3 and b = – 1

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (-1)m² + 3m + k = 0

∴   m² –  3m – k = 0 ……… (1)

One of the line is 2x + y = 0. Its slope is – 2/1 = -2

Now -2 must be one of the roots of the auxiliary equation (1),

Substituting m = -2 in equation (1)

(-2)² –  3(-2) – k = 0

∴  4 + 6 – k = 0

∴  k = 10

Example 05:

• Find k, if the one of the lines given by 6x² – 14xy  + 14ky² = 0 coincides with y = 2x
• Solution:

Given joint equation of lines is 6x² – 14xy  + 14ky² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 6, 2h = – 14 and b = 14k

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (14k)m² – 14 m + 6 = 0

∴  14km² – 14 m + 6 = 0 ……… (1)

One of the line is y = 2x. Its slope is 2

Now 2 must be one of the roots of the auxiliary equation (1),

Substituting m = 2 in equation (1)

14k(2)² – 14 (2) + 6 = 0

∴  56 k – 28 + 6 = 0

∴  56 k  = 22

∴   k  = 22/56 = 11/28

Example 06:

• Find k, if the one of the lines given by kx² – 5xy  – 6y² = 0 is 4x + 3y = 0
• Solution:

Given joint equation of lines is kx² – 5xy  – 6y² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = k, 2h = – 5 and b = – 6

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴   (- 6)m² – 5m + k = 0

∴  6m² + 5 m – k = 0……… (1)

One of the line is 4x + 3y = 0. Its slope is – 4/3

Now – 4/3  must be one of the roots of the auxiliary equation (1),

Substituting m = – 4/3 in equation (1)

6(- 4/3)² + 5 (- 4/3) – k = 0

∴  6(16/9) + 5 (- 4/3) – k = 0

Multiplying both sides of equation by 3

∴  32 – 20 – 3k = 0

– 3k = – 12

∴   k  = 4

Example 07:

• Find k, if the one of the lines given b 3x² + kxy  + 2y² = 0 is 2x + y = 0
• Solution:

Given joint equation of lines is 3x² + kxy  + 2y² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 3, 2h = k and b = 2

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴  2m² + k m + 3 = 0 ……… (1)

One of the line is 2x + y = 0. Its slope is – 2/1 = -2

Now – 2  must be one of the roots of the auxiliary equation (1),

Substituting m = – 2 in equation (1)

2(-2)² + k (-2) + 3 = 0

∴  8 – 2k + 3 = 0

∴  – 2k = – 11

∴   k  = 11/2

Example 08:

• Find the value of ‘k’ if one of the line given by 4x² + kxy – y² = 0 is 2x + y = 0.
• Solution:

Given joint equation of lines is 4x² + kxy – y² = 0

It is in the form ax² + 2hxy  + by² = 0.

a = 4, 2h = k and b = -1

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴  (-1)m² + k m + 4 = 0

∴  m² – k m – 4 = 0  ……… (1)

One of the line is 2x + y = 0. Its slope is – 2/1 = -2

Now – 2  must be one of the roots of the auxiliary equation (1),

Substituting m = – 2 in equation (1)

(-2)² – k (-2) – 4 = 0

∴  4 + 2k  – 4 = 0

∴  2k = 0

∴   k  = 0

Example 09:

• Find the value of ‘a’ if the line given by ax² + xy – 3y² = 0 is perpendicular to 3x – 5y – 1 = 0.
• Solution:

Given joint equation of lines is ax² + xy – 3y² = 0

It is in the form Ax² + 2Hxy  + By² = 0.

A = a, 2H = 1 and B = – 3

The auxiliary equation of given line is of the form

Bm² + 2Hm + A = 0

∴  (-3)m² + 1 m + a = 0

∴  3m² – m – a = 0  ……… (1)

Given line is 3x – 5y – 1 = 0. slope of this line is – 3/-5 = 3/5

One of the line is perpendicular to 3x – 5y – 1 = 0. Hence its slope = – 5/3

Now – 5/3  must be one of the roots of the auxiliary equation (1),

Substituting m = – 5/3 in equation (1)

∴  3(- 5/3)² – (- 5/3) – a = 0

∴  3(25/9) + (5/3) – a = 0

Multiplying both sides of equation by 3

∴  25 + 5 – 3a = 0

∴   – 3a = – 30

∴   a  = 10

Example 10:

• Find the value of ‘k’ if the line given by 2x² – 5xy + ky²= 0 is perpendicular to x – 2y = 8.
• Solution:

Given joint equation of lines is 2x² – 5xy + ky²= 0

It is in the form ax² + 2hxy  + by² = 0.

a = 2, 2h = – 5 and b = k

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴ km² – 5m + 2 = 0  ……… (1)

Given line is x – 2y = 8. slope of this line is – 1/-2 = 1/2

One of the line is perpendicular to x – 2y = 8. Hence its slope = – 2

Now – 2  must be one of the roots of the auxiliary equation (1),

Substituting m = – 2 in equation (1)

∴  k(-2)² – 5(-2) + 2 = 0

∴  4 k + 10 + 2 = 0

∴   4k = – 12

∴   k  = – 3

Example 11:

• Find the value of ‘k’ if the line given b 3x² – kxy + 5y²= 0 is perpendicular to 5x + 3y = 0.
• Solution:

Given joint equation of lines is 3x² – kxy + 5y²= 0

It is in the form ax² + 2hxy  + by² = 0.

a = 3, 2h = – k and b = 5

The auxiliary equation of given line is of the form

bm² + 2hm + a = 0

∴  5m² – km + 3 = 0  ……… (1)

Given line is 5x + 3y = 0. slope of this line is – 5/3

One of the line is perpendicular to 5x + 3y = 0. Hence its slope =  3/5

Now  3/5  must be one of the roots of the auxiliary equation (1),

Substituting m =  3/5 in equation (1)

∴ 5(3/5)² – k(3/5) + 3 = 0

∴ 5(9/25) – k(3/5) + 3 = 0

Multiplying both sides of equation by 5

∴  9 – 3k + 15 = 0

∴   – 3k = – 24

∴   k  = 8

Science > Mathematics > Pair of Straight Lines > Use of Auxiliary Equation of Pair of Lines

The post Use of Auxiliary Equation of Pair of Lines appeared first on The Fact Factor.

In this article, we shall study to find a joint equation of lines perpendicular to another pair of lines.

Note: If m₁ and m₂ are the slopes of the two lines represented by joint equation  ax² + 2hxy  + by² = 0. Then m₁. m₂ = a/b ; m₁ + m₂ = -2h/b

Algorithm:

1. Write given joint equation.
2. Compare with the standard equation.
3. Find values of a, h and b.
4. let m₁ and  m₂ be the slopes of the lines represented by the given joint equation.
5. Find values of m₁ + m₂ and m₁.m₂
6. Required lines are perpendicular to given lines. Hence slopes of the required lines are – 1/m1 and – 1/m2.
7. Write equations of required lines in the form u = 0 and v = 0.
8. Find u.v = 0
9. Simplify and write the joint equation of the line in standard form

Example 01:

• Find the joint equation of pair of lines through the origin and perpendicular to the line pair 5x² – 8xy + 3y² = 0.
• Solution:

Given joint equation is 5x² – 8xy + 3y² = 0.

Comparing with ax² + 2hxy  + by² = 0

a = 5, 2h = -8, and b = 3.

Now let m₁ and  m₂ be the slopes of the lines represented by  given joint equation.

m₁ + m₂ = -2h/b = -(-8)/3 = 8/3

and m₁. m₂ = a/b = 5/3

Now since the required lines are  perpendicular to the lines represented by given joint equation,

their slopes are – 1/m₁ and – 1/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   m₂y = – x

∴   x + m₂y = 0 …. (2)

combined equation is

(x + m₁y)( x + m₂y) = 0

∴   x ( x + m₂y) + m₁y ( x + m₂y) = 0

∴   x² + m₂xy + m₁xy + m₁m₂y² = 0

∴   x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴   x² + (8/3)xy + (5/3)y² = 0

Multiplying both sides by 3

3x² + 8xy + 5y² = 0

This is the required joint equation of pair of lines.

Example 02:

• Find the joint equation of pair of lines through origin and perpendicular to the line pair x² – xy + 2y² = 0
• Solution: 

Given joint equation is x² – xy + 2y² = 0

Comparing with ax² + 2hxy  + by² = 0

a = 1, 2h = -1 and b = 2.

Now let m₁ and  m₂ be the slopes of the lines represented by  given joint equation.

m₁ + m₂ = -2h/b = – (-1)/2 = 1/2

and m₁. m₂ = a/b = 1/2

Now since the required lines are  perpendicular to the lines represented by given joint equation,

their slopes are – 1/m₁ and – 1/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   m₂y = – x

∴   x + m₂y = 0 …. (2)

combined equation is

(x + m₁y)( x + m₂y) = 0

∴   x ( x + m₂y) + m₁y ( x + m₂y) = 0

∴   x² + m₂xy + m₁xy + m₁m₂y² = 0

∴   x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴   x² + (1/2)xy + (1/2)y² = 0

Multiplying both sides by 2

2x² + xy + y² = 0

This is the required joint equation of pair of lines.

Example 03:

• Find the joint equation of pair of lines through origin and perpendicular to the line pair  ax² + 2hxy + by² = 0 
• Solution:

Given joint equation is ax² + 2hxy + by² = 0

Now let m₁ and  m₂ be the slopes of the lines represented by  given joint equation.

m₁ + m₂ = -2h/b and m₁. m₂ = a/b

Now since the required lines are  perpendicular to the lines represented by given joint equation,

their slopes are – 1/m₁ and – 1/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   m₂y = – x

∴   x + m₂y = 0 …. (2)

combined equation is

(x + m₁y)( x + m₂y) = 0

∴   x ( x + m₂y) + m₁y ( x + m₂y) = 0

∴   x² + m₂xy + m₁xy + m₁m₂y² = 0

∴   x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴   x² + (-2h/b)xy + (a/b)y² = 0

Multiplying both sides by b

bx² – 2hxy + ay² = 0

This is the required joint equation of pair of lines.

Note:

• This result can be directly used in competitive exams
• To find the combined equation of the pair of lines through origin and perpendicular to the line pair 5x² – 8xy + 3y² = 0.

Here a = 5, 2h = -8 and b = 3

Hence answer is 3x² + 8xy + 5y² = 0

Example 04:

• Find the joint equation of pair of lines through origin and perpendicular to the line pair 2x² – 8xy + 3y² = 0.
• Solution: 

Given joint equation is 2x² – 8xy + 3y² = 0

Comparing with ax² + 2hxy  + by² = 0

a = 2, 2h = – 8 and b = 3.

Now let m₁ and  m₂ be the slopes of the lines represented by  given joint equation.

m₁ + m₂ = -2h/b = – (-8)/3 = 8/3

and m₁. m₂ = a/b = 2/3

Now since the required lines are  perpendicular to the lines represented by given joint equation,

their slopes are – 1/m₁ and – 1/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   m₂y = – x

∴   x + m₂y = 0 …. (2)

combined equation is

(x + m₁y)( x + m₂y) = 0

∴   x ( x + m₂y) + m₁y ( x + m₂y) = 0

∴   x² + m₂xy + m₁xy + m₁m₂y² = 0

∴   x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴   x² + (8/3)xy + (2/3)y² = 0

Multiplying both sides by 3

3x² + 8xy + 2y² = 0

This is the required joint equation of pair of lines.

Example 05:

• Find the joint equation of pair of lines through origin and perpendicular to the line pair 5x² + 2xy – 3y² = 0.
• Solution: 

Given joint equation is 2x² – 8xy + 3y² = 0

Comparing with ax² + 2hxy  + by² = 0

a = 5, 2h = 2 and b = -3.

Now let m₁ and  m₂ be the slopes of the lines represented by  given joint equation.

m₁ + m₂ = -2h/b = – (2)/-3 = 2/3

and m₁. m₂ = a/b = 5/-3 = – 5/3

Now since the required lines are  perpendicular to the lines represented by given joint equation,

their slopes are – 1/m₁ and – 1/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   m₂y = – x

∴   x + m₂y = 0 …. (2)

combined equation is

(x + m₁y)( x + m₂y) = 0

∴   x ( x + m₂y) + m₁y ( x + m₂y) = 0

∴   x² + m₂xy + m₁xy + m₁m₂y² = 0

∴   x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴   x² + (2/3)xy + (-5/3)y² = 0

Multiplying both sides by 3

3x² + 2xy – 5y² = 0

This is the required joint equation of pair of lines.

Example 06:

• Find the joint equation of pair of lines through origin and perpendicular to the line pair  x² + 4xy – 5y² = 0.
• Solution: 

Given joint equation is 2x² – 8xy + 3y² = 0

Comparing with ax² + 2hxy  + by² = 0

a = 1, 2h = 4 and b = -5.

Now let m₁ and  m₂ be the slopes of the lines represented by  given joint equation.

m₁ + m₂ = -2h/b = – (4)/-5 = 4/5

and m₁. m₂ = a/b = 1/-5 = – 1/5

Now since the required lines are  perpendicular to the lines represented by given joint equation,

their slopes are – 1/m₁ and – 1/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   m₂y = – x

∴   x + m₂y = 0 …. (2)

combined equation is

(x + m₁y)( x + m₂y) = 0

∴   x ( x + m₂y) + m₁y ( x + m₂y) = 0

∴   x² + m₂xy + m₁xy + m₁m₂y² = 0

∴   x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴   x² + (4/5)xy + (-1/5)y² = 0

Multiplying both sides by 5

5x² + 4xy – y² = 0

This is the required joint equation.

Example 07:

• Find the joint equation of pair of lines through origin and perpendicular to the line pair 2x² – 3xy – 9y² = 0.
• Solution: 

Given joint equation is 2x² – 8xy + 3y² = 0

Comparing with ax² + 2hxy  + by² = 0

a = 2, 2h = -3 and b = – 9.

Now let m₁ and  m₂ be the slopes of the lines represented by  given joint equation.

m₁ + m₂ = -2h/b = – (-3)/-9 = -1/3

and m₁. m₂ = a/b = 2/-9 = – 2/9

Now since the required lines are  perpendicular to the lines represented by given joint equation,

their slopes are – 1/m₁ and – 1/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   m₂y = – x

∴   x + m₂y = 0 …. (2)

combined equation is

(x + m₁y)( x + m₂y) = 0

∴   x ( x + m₂y) + m₁y ( x + m₂y) = 0

∴   x² + m₂xy + m₁xy + m₁m₂y² = 0

∴   x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴   x² + (-1/3)xy + (-2/9)y² = 0

Multiplying both sides by 9

9x² – 3xy – 2y² = 0

This is the required joint equation of pair of lines.

Example 08:

• Find the joint equation of pair of lines through origin and perpendicular to the line pair  x² + xy – y² = 0.
• Solution: 

Given joint equation is 2x² – 8xy + 3y² = 0

Comparing with ax² + 2hxy  + by² = 0

a = 1, 2h = 1 and b = -1.

Now let m₁ and  m₂ be the slopes of the lines represented by  given joint equation.

m₁ + m₂ = -2h/b = – 1/-1 = 1

and m₁. m₂ = a/b = 1/-1 =  -1

Now since the required lines are  perpendicular to the lines represented by given joint equation,

their slopes are – 1/m₁ and – 1/m₂ 

Equation of first required line is

y =  – 1/m₁ x

∴   m₁y = – x

∴   x + m₁y = 0 ……. (1)

and similarly equation second required line is

y =  – 1/m₂ x

∴   m₂y = – x

∴   x + m₂y = 0 …. (2)

combined equation is

(x + m₁y)( x + m₂y) = 0

∴   x ( x + m₂y) + m₁y ( x + m₂y) = 0

∴   x² + m₂xy + m₁xy + m₁m₂y² = 0

∴   x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴   x² + (1)xy + (- 1)y² = 0

x² + xy – y² = 0

This is the required joint equation of pair of lines.

Science > Mathematics > Pair of Straight Lines > Joint Equation of Pair of Lines Perpendicular To Another Pair of Lines

The post Joint Equation of Pair of Lines Perpendicular To Another Pair of Lines appeared first on The Fact Factor.

In this article, we shall study to predict the nature of lines using the joint equations of lines.

Notes:

If ax² + 2hxy + by²= 0 is a joint equation of lines then the lines represented by joint equation are

• Real if and only if h² – ab ≥ 0
• Real and distinct if and only if h² – ab > 0
• Real and coincident if and only if h² – ab = 0
• Imaginary and can’t be drawn if and only if h² – ab < 0

Algorithm:

1. Write the joint equation of lines in standard form.
2. Compare with standard ax² + 2hxy + by² = 0. Find values of a, h and b.
3. Find the value of the quantity h² – ab
4. Decide the nature of the line using the notes given above.

Example 01:

• Determine the nature of lines represented by the joint equation x² + 2xy + y² = 0
• Solution:

Given joint equation is x² + 2xy + y² = 0

Comparing with ax² + 2hxy + by² = 0

a = 1, 2h = 2, h = 1, b = 1

Now, h² – ab = (1)² – (1)(1) = 1 – 1 = 0

Here h² – ab = 0, hence the lines are real and coincident.

Example 02:

• Determine the nature of lines represented by the joint equation x² + 2xy – y² = 0
• Solution:

Given joint equation is x² + 2xy + y² = 0

Comparing with ax² + 2hxy + by² = 0

a = 1, 2h = 2, h = 1, b = -1

Now, h² – ab = (1)² – (1)(-1) = 1 + 1 = 2

Here h² – ab > 0, hence the lines are real and distinct.

Example 03:

• Determine the nature of lines represented by the joint equation 4x² + 4xy + y² = 0
• Solution:

Given joint equation is 4x² + 4xy + y² = 0

Comparing with ax² + 2hxy + by² = 0

a = 4, 2h = 4, h = 2, b =  1

Now, h² – ab = (2)² – (4)(1) = 4 – 4 = 0

Here h² – ab = 0, hence the lines are real and coincident.

Example 04:

• Determine the nature of lines represented by the joint equation x² – y² = 0
• Solution:

Given joint equation is x² + 0xy – y² = 0

Comparing with ax² + 2hxy + by² = 0

a = 1, 2h = 0, h = 0, b = – 1

Now, h² – ab = (0)² – (1)(-1) = 0 + 1 = 1

Here h² – ab > 0, hence the lines are real and distinct.

Example 05:

• Determine the nature of lines represented by the joint equation x² + 7xy + 2y² = 0

Solution:

Given joint equation is x² + 7xy + 2y² = 0

Comparing with ax²+ 2hxy + by² = 0

a = 1, 2h = 7, h = 7/2, b =  2

Now, h² – ab = (7/2)² – (1)(2) = 41/4 – 2 = 41/4

Here h² – ab > 0, hence the lines are real and distinct.

Example 06:

• Determine the nature of lines represented by the joint equation xy  = 0
• Solution:

Given joint equation is xy  = 0

Comparing with ax² + 2hxy + by² = 0

a = 0, 2h = 1, h = 1/2, b =  0

Now, h² – ab = (1/2)² – (0)(0) =1/4

Here h² – ab > 0, hence the lines are real and distinct.

Example 07:

Example – 7:  

• Determine the nature of lines represented by the joint equation x² + 2xy + 2y² = 0
• Solution:

Given joint equation is x² + 2xy + 2y² = 0

Comparing with ax² + 2hxy + by² = 0

a = 1, 2h = 2, h = 1, b =  2

Now, h² – ab = (1)² – (1)(2) = 1 – 2 = – 1

Here h² – ab < 0, hence the lines are imaginary and can’t be drawn.

Example 08:

• Determine the nature of lines represented by the joint equation px² + 2qxy – py² = 0.
• Solution:

Given joint equation is px² + 2qxy – py² = 0

Comparing with ax² + 2hxy + by² = 0

a = p, 2h = 2q, h = q, b = – p

Now, h² – ab = (q)² – (p)(-p) = q² + p²

as p and q are real numbers, there squares are always positive. Thus sum of their squares is always positive. Here h² – ab > 0, hence the lines are real and distinct.

Example 09:

• Determine the nature of lines represented by the joint equation px² – qy² = 0
• Solution:

Given joint equation is px² + 0xy – qy² = 0

Comparing with ax² + 2hxy + by²= 0

a = p, 2h = 0, h = 0, b = – q

Now, h² – ab = (0)² – (p)(-q) = 0 + pq = pq

• If p and q are both positive, then the product pq is positive, then h² – ab > 0, hence the lines are real and distinct.
• If p and q are both negative, then the product pq is positive, then h² – ab > 0, hence the lines are real and distinct.
• If p and q have opposite signs then the product pq is negative,  then h²– ab < 0, hence the lines are imaginary and can’t be drawn.
• If p = q = 0, then pq = 0. Here h² – ab = 0, hence the lines are real and coincident.

Nature of Lines is Given. To Find the Value of Constant

Algorithm:

1. Write the joint equation of lines in standard form.
2. Compare with standard ax² + 2hxy + by² = 0. Find values of a, h and b.
3. Find the value of the quantity h² -ab.
4. Use the conditions depending upon the nature of the line use notes given at the top and find the value of the constant.

Example 10:

• If the lines represented by equation x² + 2hxy + 4y² = 0  are real and coincident, find h.
• Solution:

Given joint equation is x² + 2hxy + 4y² = 0

Comparing with Ax² + 2Hxy + By² = 0

A = 1, 2H = 2h, H = h, B =  4

Now, lines are real and coincident

H² – AB = 0

h² – (1)(4) = 0

h² – 4 = 0

h² = 4

h = ± 2

Example 11:

• If the lines represented by equation px² + 6xy + 9y² = 0  are real and distinct, find p.
• Solution:

Given joint equation is px2 + 6xy + 9y2  = 0

Comparing with ax² + 2hxy + by² = 0

a = p, 2h = 6, h = 3, b =  9

Now, lines are real and distinct

h² – ab > 0

32 – (p)(9) > 0

9 – 9p > 0

-9p > -9

∴ p < 1

∴ p ∈ (- ∞, 1)

Example 12:

• If the lines represented by equation px² + 4xy + 4y² = 0  are real and distinct, find p.
• Solution:

Given joint equation is px² + 4xy + 4y² = 0

Comparing with ax² + 2hxy + by² = 0

a = p, 2h = 4, h = 2, b =  4

Now, lines are real and distinct

h² – ab > 0

2² – (p)(4) > 0

4 – 4p > 0

-4p > -4

∴ p < 1

∴ p ∈ (- ∞, 1)

Example 13:

• If the lines represented by equation 3x² + 2hxy + 3y² = 0  are real , find h.
• Solution:

Given joint equation is 3x² + 2hxy + 3y² = 0

Comparing with Ax² + 2Hxy + By² = 0

A = 3, 2H = 2h, H = h, B =  3

Now, lines are real and coincident

H² – AB  = 0

h² – (3)(3) =  0

h² – 9  = 0

∴ h² =  9

∴  h  = ± 3

Science > Mathematics > Pair of Straight Lines > Nature of Lines Represented by Joint Equation

The post Nature of Lines Represented by Joint Equation appeared first on The Fact Factor.

In this article, we shall study to find separate equations of lines from a combined or joint equation of pair of lines by auxiliary equation method.

Algorithm:

1. Check if lines exist. use the same method used in the case to find the nature of lines. Proceed further if lines exist.
2. Divide both sides of the equation by x².
3. Simplify the equation and substitute y/x = m in it.
4. Find two roots m₁ and m₂ of quadratic equation in m
5. Find separate equations of lines by y = m₁ x and y = m₂x
6. Note that this method is applicable to any problem.

Example 01:

• Obtain the separate equations of the lines represented by  11x² + 8xy + y² = 0
• Solution:

The given joint equation is  11x² + 8xy + y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in equation (1) we get

11 + 8m + m² = 0

m² + 8m + 11 = 0

This is a quadratic equation in m and has two roots say m₁ and m₂

which gives slopes of the two lines represented by the joint equation.

a = 1, b = 8, c = 11

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (-4 + √5) x

∴  y = – (4 – √5) x

 ∴  (4 – √5) x  + y = 0

The equation of the second line is

y = m₂x  i.e.  i.e.

∴  y = (-4 – √5) x

∴  y = – (4 + √5) x

 ∴  (4 + √5) x  + y = 0

Ans: The separate equations of the lines are (4 – √5) x  + y = 0 and (4 + √5) x  + y = 0

Example 02:

• Obtain separate equations of lines represented by joint equation x² – 4xy + y² = 0.
• Solution:

The given joint equation is  x² – 4xy + y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 – 4m + m² = 0

∴  m² – 4m + 1= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = -4, c = 1

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (2 + √3) x

 ∴  (2 + √3) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (2 – √3) x

 ∴  (2 – √3) x  – y = 0

Ans: The separate equations of the lines are  (2 + √3) x  – y = 0 and (2 – √3) x  – y = 0

Example 03:

• Obtain separate equations of lines represented by joint equation 22x² – 10xy + y² = 0.

Solution:

The given joint equation is 22x² – 10xy + y² = 0.

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

22 – 10m + m² = 0

∴  m² – 10m + 22= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 10, c = 22

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (5 + √3) x

 ∴  (5 + √3) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (5 – √3) x

 ∴  (5 – √3) x  – y = 0

Ans: The separate equations of the lines are  (5 + √3) x  – y = 0 and (5 – √3) x  – y = 0

Example 04:

• Obtain separate equations of lines represented by joint equation x² + 2xy – y² = 0.
• Solution:

The given joint equation is x² + 2xy – y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 + 2m – m² = 0

∴  m² – 2m  – 1= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 2, c = -1

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (1 + √2) x

 ∴  (1 + √2) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (1 – √2) x

 ∴  (1 – √2) x  – y = 0

Ans: The separate equations of the lines are  (1 + √2) x  – y = 0 and (1 – √2) x  – y = 0

Example 05:

• Obtain separate equations of lines represented by joint equation 2x² + 2xy – y² = 0.

Solution:

The given joint equation is  2x² + 2xy – y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

2 + 2m – m² = 0

∴  m² – 2m – 2= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 2, c = -2

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (1 + √3) x

 ∴  (1 + √3) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (1 – √3) x

 ∴  (1 – √3) x  – y = 0

Ans: The separate equations of the lines are  (1 + √3) x  – y = 0 and (1 – √3) x  – y = 0

Example 06:

• Obtain separate equations of lines represented by joint equation x² + 2(tanα) xy – y² = 0.
• Solution:

The given joint equation is x² + 2(tanα) xy – y² = 0

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 + 2tanα m – m² = 0

∴  m² – 2tanα m – 1= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 2tanα, c = -1

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (tanα + secα) x

 ∴  (tanα + secα) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (tanα – secα) x

 ∴  (tanα – secα) x  – y = 0

Ans: The separate equations of the lines are  (tanα + secα) x  – y = 0 and  (tanα – secα) x  – y = 0

Example 07:

• Obtain separate equations of lines represented by joint equation x² + 2(cosecα) xy + y² = 0.
• Solution:

The given joint equation is x² + 2(cosecα) xy + y² = 0.

Dividing both sides of the equation by x²

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 + 2cosecα m + m² = 0

∴  m² + 2cosecα m + 1= 0

This is a quadratic equation in m and has two roots say m₁ and m₁

which gives slopes of the two lines represented by the joint equation.

a = 1, b = 2cosecα, c = 1

The roots are given by

The equation of the first line is

y = m₁x

∴  y = (- cosec α + cot α) x

∴  y = –  (cosec α – cot α) x

 ∴  (cosec α – cot α) x  – y = 0

The equation of the second line is

y = m₂x

∴  y = (- cosec α – cot α) x

∴  y = – (cosec α + cot α) x

 ∴  (cosec α + cot α) x  – y = 0

Ans: The separate equations of the lines are  (cosec α – cot α) x  – y = 0 and  (cosec α + cot α) x  – y = 0

Science > Mathematics > Pair of Straight Lines > Separate Equations of Lines (Auxillary Equation Method)

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