Physics | Chemistry | Biology | Mathematics |

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

**Type – IF: To Find Joint Equation of Pair of Lines, ****When Point and Two Other Lines are Given**

**Notes:**

- Slope of ax + by + c = 0 is – a/b (b ≠ 0) i.e. Slope = – coefficient of x / coefficient of y
- When two lines are parallel their slopes are equal.
- When two lines are perpendicular to each other then the product of their slopes is -1. Thus, If slope one line is m, then the slope of a line perpendicular to it is – 1/m.

**ALGORITHM:**

- Find the slopes of given line using the formula given in point 1 of notes.
- Find the slopes of required lines forming pair using relations given in points 2 and 3 of notes.
- Use slope point form y – y
_{1}= m (x – x_{1}) to find equations of given lines. If Lines are passing through origin use y = mx form. - 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 30:**

- Find the joint equation of a pair of lines through the origin such that one is parallel to x + 2y = 5 and other is perpendicular to 2x – y + 3 = 0.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

The equation of first given line is x + 2y = 5, Hence slope of this line is – 1/2

As required first line (*l*_{1}) is parallel to this line. The slope of *l*_{1} is – 1/2

The equation of the line (*l*_{1}) passing through the origin is

y = -1/2 x

∴ 2y = – x

∴ x + 2y = 0 …. (1)

Equation of second given line is 2x – y + 3 = 0. Hence slope of this line is -2/-1 = 2

As required first line (*l*_{2}) is perpendicular to this line Slope of *l*_{2} is – 1/2

The equation of line (2) passing through origin is

y =-1/2 x

∴ 2y = -x

∴ x + 2y = 0 ……. (2)

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

(x + 2y) (x + 2y) = 0

∴ x^{2} + 4xy + 4y^{2} = 0

This is the required combined equation for the pair of lines.

**Example 31:**

- Find the joint equation of a pair of lines through the origin such that one is parallel to 2x + y – 5 = 0 and other is perpendicular to 3x – 4y + 7 = 0.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

Equation of first given line is 2x + y – 5 = 0. Hence slope of this line is -2/1 = -2

As required first line (* l*_{1}) is parallel to this line. Hence slope of * l*_{1} is – 2

The equation of line (1) passing through origin is

∴ y = – 2 x

∴ 2x + y = 0 …. (1)

Equation of second given line is 3x – 4y + 7 = 0. Hence slope of this line is – 3/-4 = 3/4

As required first line (*l*_{2}) is perpendicular to this line. Hence slope of *l*_{2} is -4/3

The equation of line (*l*_{2}) passing through origin is

y =* *– 4/3 x

∴ 3y = – 4x

∴ 4x + 3y = 0 ……. (2)

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

(2x + y) (4x + 3y) = 0

∴ 8x^{2} + 6xy + 4xy + 3y^{2} = 0

∴ 8x^{2} + 10xy + 3y^{2} = 0

This is the required combined equation for the pair of lines.

**Example 32:**

- Find the joint equation of a pair of lines through the origin such that one is parallel to 3x – y = 7 and other is perpendicular to 2x + y = 8.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

Equation of first given line is 3x + y – 7 = 0. Hence slope of this line is – 3/1 = – 3

As required first line (* l*_{1}) is parallel to this line. Hence slope of * l*_{1} is – 3

The equation of line (* l*_{1}) passing through origin is

y = – 3 x

∴ 3x + y = 0 …. (1)

Equation of second given line is 2x + y – 8 = 0. Hence slope of this line is -2/1 = – 2

As required first line (*l*_{2}) is perpendicular to this line. Hence slope of *l*_{2} is – 2

The equation of line (2) passing through origin is

y = -2x

∴ 2x + y = 0 …. (2)

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

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

∴ 6x^{2} + 6xy + 2xy + y^{2} = 0

∴ 6x^{2} + 8xy + y^{2} = 0

This is the required combined equation for the pair of lines.

**Example 33: **

- Find the joint equation of pair of lines through the origin, one of which is parallel to and another is perpendicular to the line 2x + 3y – 2 = 0.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

The equation of the given line is 2x + 3y – 2 = 0.

Slope of the given line = m = -2/3

Let *l*_{1} be parallel to given line slope of *l*_{1} is -2/3

The equation of line (1) passing through origin is

y = – 2/3 x

∴ 3 y = – 2 x

∴ 2x + 3y = 0 …… (1)

Let *l*_{2} be perpendicular to given line slope of *l*_{2} is 3/2

The equation of line (*l*_{2}) passing through origin is

y = 3/2 x

∴ 2 y = 3 x

∴ 3x – 2y = 0 …. (2)

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

(2x + 3y) (3x – 2y) = 0

∴ 2x(3x – 2y) + 3y(3x – 2y) = 0

∴ 6x^{2} – 4xy + 9xy – 6y^{2} = 0

∴ 6x^{2} + 5xy – 6y^{2} = 0

This is the required combined equation for the pair of lines.

**Example 34: **

- Find the joint equation of pair of lines through the origin, one of which is parallel to and another is perpendicular to the line 5x + 3y = 7.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

The equation of the given line is 5x + 3y – 7 = 0.

Slope of the given line = m = – 5/3

Let *l*_{1} be parallel to given line slope of *l*_{1} is -5/3

The equation of line (*l*_{1}) passing through origin is

y = – 5/3 x

∴ 3 y = – 5 x

∴ 5x + 3y = 0 …………… (1)

Let *l*_{2} be perpendicular to given line slope of *l*_{2} is 3/5

The equation of line (*l*_{2}) passing through origin is

y = 3/5 x

∴ 5 y = 3 x

∴ 3x – 5y = 0 …. (2)

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

(5x + 3y) (3x – 5y) = 0

∴ 5x(3x – 5y) + 3y(3x – 5y) = 0

∴ 15x^{2} – 25xy + 9xy – 15y^{2} = 0

∴ 15x^{2} – 16xy – 15y^{2} = 0

This is the required combined equation.

**Example 35: **

- Find the joint equation of a pair of lines through the origin such that one is parallel to and other is perpendicular to x + 2y + 1857 = 0.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

The equation of the given line is x + 2y + 1857 = 0.

The slope of the given line = m = =1/2

Let *l*_{1} be parallel to given line slope of *l*_{1} is – 1/2

The equation of line (*l*_{1}) passing through origin is

y = – 1/2 x

∴ 2y = – x

∴ x + 2y = 0 …… (1)

Let *l*_{2} be perpendicular to given line slope of *l*_{2} is 2

The equation of line (*l*_{2}) passing through origin is

y = 2 x

∴ 2x – y = 0 …. (2)

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

(x + 2y) (2x – y) = 0

∴ x(2x – y) + 2y(2x – y) = 0

∴ 2x^{2} – xy + 4xy – 2y^{2} = 0

∴ 2x^{2} + 3xy – 2y^{2} = 0

This is the required combined equation.

**Example 36: **

- Find the joint equation of pair of lines through the origin, and perpendicular to the lines x + 2y = 19 and 3x + y = 18.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

The equation of the first given line is x + 2y – 19=0.

Slope of the first given line = m = – 1/2

Let *l*_{1} be perpendicular to firstgiven line. Hence slope of *l*_{1} is 2

The equation of line (*l*_{1}) passing through origin is

y = 2x

∴ 2x – y = 0 …… (1)

The equation of the second given line is 3x + y – 18 = 0

Slope of the second given line = m = -3/1 = -3

Let *l*_{2} be perpendicular to second given line. Hence slope of *l*_{2} is 1/3

The equation of line ( *l*_{2}) passing through origin is

y = 1/3** **x

∴ 3 y = x

∴ x – 3y = 0 …. (2)

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

(2x – y) (x – 3y) = 0

∴ 2x(x – 3y) – y(x – 3y) = 0

∴ 2x^{2} – 6xy – xy + 3y^{2} = 0

∴ 2x^{2} – 7xy + 3y^{2} = 0

This is the required combined equation.

**Example 37:**

- Find the joint equation of a pair of lines through the point (3, 4) and such that one is parallel to 2x + 3y + 7 = 0 and other is perpendicular to 3x – 5y = 8.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

Equation of first given line is 2x + 3y + 7 = 0

Hence slope of this line is – 2/3

As required first line (*l*_{1}) is parallel to this line

Slope of *l*_{1} is = m = -2/3

The equation of line (*l*_{1}) passing through (3, 4) ≡ (x_{1}, y_{1}) is

(y – y_{1}) = m(x – x_{1})

∴ (y – 4) = -2/3 (x – 3)

∴ 3y -12 = – 2x + 6

∴ 2x + 3y – 18 = 0 ……. (1)

Equation of second given line is 3x – 5y – 8 = 0.

Hence slope of this line is – 3/-5 = 3/5

As required first line (*l*_{2}) is perpendicular to this line

Slope of *l*_{2} is = m = -5/3

The equation of line (*l*_{2}) passing through (3, 4) ≡ (x_{1}, y_{1}) is

(y – y_{1}) = m(x – x_{1})

∴ (y – 4) = – 5/3 (x – 3)

∴ 3y -12 = – 5x + 15

∴ 5x + 3y – 27 = 0 …. (2)

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

(2x + 3y – 18) (5x + 3y – 27) = 0

∴ 2x(5x + 3y – 27) +3y(5x + 3y – 27) -18(5x + 3y – 27) = 0

∴ 10x^{2} + 6xy – 54x + 15xy + 9y^{2} – 81y – 90x -54 y + 486= 0

∴ 10x^{2} + 21xy + 9y^{2} – 144x – 135y + 486= 0

This is the required combined equation.

**Example 38:**

- Find the joint equation of a pair of lines through the point (- 1, 2) and perpendicular to lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

Equation of first given line is x + 2y + 3 = 0

Hence slope of this line is – 1/2

As required first line (*l*_{1}) is perpendicular to this line

Slope of *l*_{1} is = m = 2

The equation of line (*l*_{1}) passing through (-1, 2) ≡ (x_{1}, y_{1}) is

(y – y_{1}) = m(x – x_{1})

∴ (y – 2) = 2(x + 1)

∴ y -2 = 2x + 2

∴ 2x – y + 4 = 0 ……. (1)

Equation of second given line is 3x – 4y – 5 = 0.

Hence slope of this line is – 3/-4 = 3/4

As required first line (*l*_{2}) is perpendicular to this line

Slope of *l*_{2} is = m = -4/3

The equation of line (*l*_{2}) passing through (-1, 2) ≡ (x_{1}, y_{1}) is

∴ (y – y_{1}) = m(x – x_{1})

∴ (y – 2) = – 4/3 (x + 1)

∴ 3y – 6 = – 4x – 4

∴ 4x + 3y – 2 = 0 ……. (2)

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

( 2x – y + 4) (4x + 3y – 2) = 0

∴ 2x(4x + 3y – 2) – y(4x + 3y – 2) + 4(4x + 3y – 2) = 0

∴ 8x^{2} + 6xy – 4x – 4xy – 3y^{2} + 2y + 16x + 12 y – 8 = 0

∴ 8x^{2} + 2xy – 3y^{2} + 12x + 14y – 8= 0

This is the required combined equation.

**Example 39:**

- Find the joint equation of a pair of lines through the point (3, 2) one of which is parallel to x – 2y = 2 and other is perpendicular to y = 3.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

Equation of first given line is x – 2y – 2 = 0

Hence slope of this line is -1 /-2 = 1/2

As required first line (*l*_{1}) is parallel to this line

Slope of *l*_{1} is = m = 1/2

The equation of line (*l*_{1}) passing through (3, 2) ≡ (x_{1}, y_{1}) is

(y – y_{1}) = m(x – x_{1})

∴ (y – 2) = (x – 3)

∴ 2y – 4 = x – 3

∴ x – 2y+ 1 = 0 …. (1)

Equation of second given line is y = 3.

The equation of line (*l*_{2}) passing through (3, 2) and perpendicular to y = 3 is

x = 3

∴ x – 3 = 0 ……………. (2)

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

( x – 3) (x – 2y+ 1) = 0

∴ x(x – 2y+ 1) – 3(x – 2y+ 1) = 0

∴ x^{2} – 2xy + x – 3x + 6y – 3 = 0

∴ x^{2} – 2xy – 2x + 6y – 3= 0

This is the required combined equation.

**Example 40:**

- Find the joint equation of a pair of lines through the point (1, 2) and perpendicular to both the lines 3x + 2y – 5 = 0 and 2x – 5y + 1 = 0.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

Equation of first given line is 3x + 2y – 5 = 0

Hence slope of this line is -3/2

As required first line (*l*_{1}) is perpendicular to this line

Slope of *l*_{1} is = 2/3 = m

The equation of line (*l*_{1}) passing through (1, 2) ≡ (x_{1}, y_{1}) is

(y – y_{1}) = m(x – x_{1})

∴ (y – 2) = 2/3(x – 1)

∴ 3y – 6 = 2x – 2

∴ 2x – 3y + 4 = 0 …. (1)

Equation of second given line is 2x – 5y + 1 = 0.

Hence slope of this line is – 2/-5 = 2/5

As required first line (*l*_{2}) is perpendicular to this line

Slope of *l*_{2} is = -5/2 = m

The equation of line (*l*_{2}) passing through (3, 4) = (x_{1}, y_{1}) is

(y – y_{1}) = m(x – x_{1})

∴ (y – 2) = – 5/2(x – 1)

∴ 2y -4 = – 5x + 5

∴ 5x + 2y – 9 = 0 …. (2)

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

(2x – 3y + 4) (5x + 2y – 9) = 0

∴ x(5x + 2y – 9) – 3y(5x + 2y – 9) +4(5x + 2y – 9) = 0

∴ 10x^{2} + 4xy – 18x – 15xy – 6y^{2} + 27y + 20x + 8y – 36= 0

∴ 10x^{2} – 11xy – 6y^{2} + 2x + 35y – 36= 0

This is the required combined equation for the pair of lines.

**Example 41:**

- Find the joint equation of a pair of lines through the point (2, 3) and perpendicular to both the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0.
**Solution:**

Let* l*_{1} and *l*_{2} be the two lines whose joint equation is to be found.

Equation of first given line is 3x + 2y – 1 = 0

Hence slope of this line is – 3/2

As required first line (*l*_{1}) is perpendicular to this line

Slope of *l*_{1} is = 2/3 = m

The equation of line (*l*_{1}) passing through (1, 2) = (x_{1}, y_{1}) is

(y – y_{1}) = m(x – x_{1})

∴ (y – 3) = 2/3 (x – 2)

∴ 3y – 9 = 2x – 4

∴ 2x – 3y + 5 = 0 ……. (1)

Equation of second given line is x – 3y + 2 = 0.

Hence slope of this line is – 1/-3 = 1/3

As required first line (*l*_{2}) is perpendicular to this line

Slope of *l*_{2} is – 3 = m

The equation of line (*l*_{2}) passing through (3, 4) = (x_{1}, y_{1}) is

(y – y_{1}) = m(x – x_{1})

∴ (y – 3) = -3 (x – 2)

∴ y – 3 = – 3x + 6

∴ 3x + y – 9 = 0 …. (2)

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

(2x – 3y + 5) (3x + y – 9) = 0

∴ 2x(3x + y – 9) – 3y(3x + y – 9) +5(3x + y – 9) = 0

∴ 6x^{2} + 2xy – 18x – 9xy – 3y^{2} + 27y + 15x + 5y – 45= 0

∴ 6x^{2} – 7xy – 3y^{2} – 3x + 32y – 45= 0

This is the required combined equation.

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

Physics | Chemistry | Biology | Mathematics |