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Science > Mathematics > Pair of Straight Lines > You are Here |

**Type – IC: ****When Lines are Forming Equilateral Triangle With x = k or y = k are Given**

**ALGORITHM:**

- Find angle made by the lines with the positive direction of x-axis and calculate slopes m1 and m2 of the two lines.
- Use y = mx form to find equations of the two lines.
- Write equations of lines in the form u = 0 and v = 0.
- Find u.v = 0.
- Simplify the L.H.S. of the joint equation.

**Example – 17:**

- Find the joint equation of the straight lines through the origin which forms the equilateral triangle with the straight line x = 1.
**Solution:**

** **

** **Let OA and OB be two lines such that triangle OAB is an equilateral triangle.

Equation of AB is x = 1 and AB, is parallel to the y-axis.

Now triangle OAB is an equilateral triangle,

by symmetry, m ∠ MOA = m ∠MOB = 30°

Slope of OA = m_{1} = tan(-30°) = – tan 30° = – 1/√3 and

Slope of OB = m_{2} = tan( 30°) = 1/√3

Equation of OA is passing through the origin, is

y = m_{1}x. i.e. y = – 1/√3 x

∴ √3 y = – x

∴ x + √3 y = 0 …….. (1)

As OB is also passing through the origin, the form of the equation of OB is

y = m_{2}x. i.e. y = 1/√3 x

∴ √3 y = x

∴ x – √3 y = 0 …….. (2)

Hence, the combined equation of the pair OA and OB is

(x + √3 y)(x – √3 y ) = 0

∴ x ^{2} – 3y ^{2} = 0.

This is the required combined equation.

**Note:**

- The joint equation of the straight lines through the origin which forms the equilateral triangle with any straight line x = k is always x
^{2}– 3y^{2}= 0.

**Example 18 :**

- Find the joint equation of the straight lines through the origin which forms the equilateral triangle with the straight line y = 2.
**Solution:**

Let OA and OB be two lines such that the triangle OAB is an equilateral triangle.

The equation of AB is y = 2 and AB is parallel to the x-axis.

Now since triangle OAB is an equilateral triangle,

m ∠ XOA = 60° and m ∠ XOB = 120°

∴ Slope of OA = tan 60° **= √3 **and

∴ slope of OB = tan 120° **= **– √3** **

As OA is passing through the origin, the form of equation of OA is

y = m_{1}x. i.e. y = √3 x

∴ √3 x – y = 0 ………. (1)

As OB is also passing through the origin, the form of equation of OB is y = mx.

y = m_{2}x. i.e. y = – √3 x

∴ √3 x + y = 0 ………. (2)

Hence their combined equation is

( √3 x – y)(** **√3** x + y**) = 0

∴ 3x^{2} – y^{2} = 0.

**Note:**

- The joint equation of the straight lines through the origin which forms the equilateral triangle with any straight line y = k is always 3x
^{2}– y^{2}= 0.

**Type – ID: ****When Slopes of Lines are Given**

**ALGORITHM :**

- Locate slopes m1 and m2 of the two lines.
- Use y = mx form to find equations of the two lines.
- Write equations of lines in the form u = 0 and v = 0. Find u.v = 0.
- Simplify the L.H.S. of the joint equation.

**Example 19:**

- Find the joint equation of a pair of lines through the origin having slopes 1 and 3.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines. Slope of *l*_{1} is 1 and that of *l*_{2} is 3

Therefore the equation of line (*l*_{1}) passing through origin and having slope 1 is

y = 1 (x)

∴ x – y = 0 ……. (1)

Similarly the equation of line (*l*_{2}) passing through origin and having slope 3 is

y = 3 (x)

∴ 3 x – y = 0 …. (2)

From (1) and (2) the required combined equation is

(x – y) (3x – y) = 0

∴ x(3x – y) – y(3x – y) = 0

∴ 3x^{2} – xy – 3xy + y^{2} = 0

∴ 3x^{2} – 4xy + y^{2} = 0

This is the required cobined equation.

**Example 20:**

- Find the joint equation of a pair of lines through the origin having slopes 4 and -1.

**Solution:**

Let* l*_{1} and *l*_{2} be the two lines. Slope of *l*_{1} is 4 and that of *l*_{2} is – 1

Therefore the equation of line (*l*_{1}) passing through origin and having slope 4 is

y = 4 (x)

∴ 4x – y = 0 ……. (1)

Similarly the equation of line (*l*_{2}) passing through origin and having slope -1 is

y = -1 (x)

∴ x + y = 0 …. (2)

From (1) and (2) the required combined equation is

(4x – y) (x + y) = 0

∴ 4x(x + y) – y(x + y) = 0

∴ 4x^{2} + 4xy – xy + y^{2} = 0

∴ 4x^{2} + 3xy + y^{2} = 0

This is the required cobined equation.

**Example 21:**

- Find the joint equation of a pair of lines through the origin having slopes 2 and – 1/2.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines. Slope of *l*_{1} is 2 and that of *l*_{2} is – 1/2

Therefore the equation of line ( *l*_{1}) passing through origin and having slope 2 is

y = 2 (x)

∴ 2x – y = 0 ……. (1)

Similarly the equation of line (*l*_{2}) passing through origin and having slope is

y = -1/2 (x)

∴ 2y = – x

∴ x + 2y = 0 ………. (2)

From (1) and (2) the required combined equation is

(2x – y) (x + 2y) = 0

∴ 2x(x + 2y) – y(x + 2y) = 0

∴ 2x^{2} + 4xy – xy – 2y^{2} = 0

∴ 2x^{2} + 3xy – 2y^{2} = 0

This is the required cobined equation.

**Example 22:**

- Find the joint equation of a pair of lines through the origin having slopes 1/3 and -1/2.

**Solution:**

Let* l*_{1} and *l*_{2} be the two lines. Slope of *l*_{1} is 1/3 and that of *l*_{2} is – 1/2

Therefore the equation of line (*l*_{1}) passing through origin and having slope is

y = 1/3 (x)

∴ 3y = x

∴ x – 3y = 0 ……. (1)

Similarly the equation of line (*l*_{2}) passing through origin and having slope is

y = (x)

∴ 2y = -x

∴ x + 2y = 0 ……. (2)

From (1) and (2) the required combined equation is

(x – 3y) (x + 2y) = 0

∴ x(x + 2y) – 3y(x + 2y) = 0

∴ x^{2} + 2xy – 3xy – 6y^{2} = 0

∴ x^{2} – xy = 6y^{2} = 0

This is the required cobined equation.

**Example 23:**

- Find the joint equation of a pair of lines through the origin having slopes 1 + √3 and 1 – √3.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines. Slope of *l*_{1} is 1 + √3 and that of *l*_{2} is 1 – √3

Therefore the equation of line (*l*_{1}) passing through origin and having slope is

y = (1 + √3)x

∴ (1 + √3)x – y = 0 …. (1)

Similarly, the equation of the line (*l*_{2}) passing through the origin and having slope is

y = ( 1 – √3)x

∴ ( 1 – √3)x – y = 0 …. (2)

From (1) and (2) the required combined equation is

This is the required combined equation.

**Type – IE: ****When Point and Lines Parallel to Co-ordinate Axes are Given**

**ALGORITHM :**

- When line passes through (h, k) and is parallel to co-ordinate axes are given. Equations of lines are x = h and y = k
- Write equations of lines in the form u = 0 and v = 0.
- Find u.v = 0.
- Simplify the L.H.S. of the joint equation.

**Example 24:**

- Find the joint equation of a pair of lines which passes through (1, 2) and parallel to co-ordinate axes.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines.

Equation of *l*_{1} passing through (1, 2) and parallel to y-axis is x = 1

∴ x – 1 = 0 …………. (1)

Equation of *l*_{2} passing through (1, 2) and parallel to x-axis is y = 2

∴ y – 2 = 0 …………. (2)

The required joint equation is

(x – 1)(y – 2) = 0

∴ xy – 2x – y + 2 = 0

This is the required cobined equation.

**Example 25:**

- Find the joint equation of a pair of lines which passes through (-3, 4) and parallel to co-ordinate axes.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines.

Equation of *l*_{1} passing through (-3, 4) and parallel to y – axis is x = -3

∴ x + 3 = 0 …………. (1)

Equation of *l*_{2} passing through (-3, 4) and parallel to x – axis is y = 4

∴ y – 4 = 0 …………. (2)

The required joint equation is

(x + 3)(y – 4) = 0

∴ xy – 4x + 3y – 12 = 0

This is the required cobined equation.

**Example 27:**

- Find the joint equation of a pair of lines which passes through (3, 2) and parallel to the lines x = 2 and y = 3.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines.

Equation of *l*_{1} passing through (3, 2) and parallel to x = 2 is x = 3

∴ x – 3 = 0 …………. (1)

Equation of *l*_{2} passing through (3, 2) and parallel to y = 3 is y = 2

∴ y – 2 = 0 …………. (2)

The required joint equation is

(x – 3)(y – 2) = 0

∴ xy – 2x – 3y + 6 = 0

This is the required cobined equation.

**Example 27:**

- Find the joint equation of a pair of lines which passes through (2, 3) and parallel to co-ordinate axes.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines.

Equation of *l*_{1} passing through (1, 2) and parallel to y – axis is x = 2

∴ x – 2 = 0 …………. (1)

Equation of *l*_{2} passing through (1, 2) and parallel to x – axis is y = 3

∴ y – 3 = 0 …………. (2)

The required joint equation is

(x – 2)(y – 3) = 0

∴ xy – 3x – 2y + 6 = 0

This is the required cobined equation.

**Example 28:**

- Find the joint equation of a pair of lines which are at a distance of 9 units from the y-axis and parallel to it.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines.

Equation of *l*_{1} which is at a distance of 9 from y – axis and parallel to it is x = – 9 i.e. x + 9 = 0 …. (1)

Equation of *l*_{2} which is at a distance of 9 from y – axis and parallel to it is x = 9 i.e. x – 9 = 0 …… (1)

The required joint equation is

(x + 9)(x – 9) = 0

∴ x^{2} – 81 = 0

This is the required cobined equation.

**Example – 29:**

- Find the joint equation of a pair of lines which are at a distance of 5 units from the x-axis and parallel to it.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines.

Equation of *l*_{1} which is at a distance of 5 from x – axis and parallel to it is y = 5i.e. y – 5 = 0 ……… (1)

Equation of *l*_{2} which is at a distance of 5 from x – axis and parallel to it is y = -5 i.e. y + 5 = 0 …… (2)

The required joint equation is

(y – 5)(y + 5) = 0

∴ y^{2} – 25 = 0

This is the required cobined equation.

Science > Mathematics > Pair of Straight Lines > You are Here |

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