# Combined Equation of Pair of Lines – 03

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#### Notes:

1. Slope of ax + by + c = 0 is – a/b (b ≠ 0) i.e. Slope = – coefficient of x / coefficient of y
2. When two lines are parallel their slopes are equal.
3. 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:

1. Find the slopes of given line using the formula given in point 1 of notes.
2. Find the slopes of required lines forming pair using relations given in points 2 and 3 of notes.
3. Use slope point form  y – y1 =  m (x – x1 ) to find equations of given lines. If Lines are passing through origin use y = mx form.
4. Write equations of lines in the form u = 0 and  v = 0.
5. Find u.v = 0.
6. 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 l1 and l2  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 (l1) is parallel to this line. The slope of l1 is  – 1/2

The equation of the line (l1) 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 (l2) is perpendicular to this line Slope of l2 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

∴   x2 + 4xy + 4y2 = 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 l1 and l2  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 ( l1) is parallel to this line. Hence slope of l1 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 (l2) is perpendicular to this line. Hence slope of l2 is -4/3

The equation of line (l2) 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

∴   8x2 + 6xy + 4xy + 3y2 = 0

∴   8x2 + 10xy + 3y2 = 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 l1 and l2  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 ( l1) is parallel to this line. Hence slope of l1 is – 3

The equation of line ( l1) 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 (l2) is perpendicular to this line. Hence slope of l2 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

∴   6x2 + 6xy + 2xy + y2 = 0

∴   6x2 + 8xy + y2 = 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 l1 and l2  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  l1 be parallel to given line slope of l1  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  l2 be perpendicular to given line slope of l2  is 3/2

The equation of line (l2) 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

∴    6x2 – 4xy + 9xy – 6y2 = 0

∴    6x2 + 5xy – 6y2 =  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 l1 and l2  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  l1 be parallel to given line slope of l1  is -5/3

The equation of line (l1) passing through origin is

y =  – 5/3 x

∴    3 y = – 5 x

∴    5x + 3y = 0 …………… (1)

Let  l2 be perpendicular to given line slope of l2  is 3/5

The equation of line (l2) 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

∴    15x2 – 25xy + 9xy – 15y2 = 0

∴    15x2 – 16xy – 15y2 =  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 l1 and l2  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 l1 be parallel to given line slope of l1  is – 1/2

The equation of line (l1) passing through origin is

y =  – 1/2 x

∴    2y = – x

∴    x + 2y = 0 …… (1)

Let  l2 be perpendicular to given line slope of l2  is 2

The equation of line (l2) 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

∴    2x2 – xy + 4xy – 2y2 = 0

∴    2x2 + 3xy – 2y2 =  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 l1 and l2  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  l1 be perpendicular to firstgiven line. Hence  slope of l1  is  2

The equation of line (l1) 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 l2  be perpendicular to second given line. Hence slope of l2  is 1/3

The equation of line ( l2) 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

∴    2x2 – 6xy – xy + 3y2 = 0

∴    2x2 – 7xy + 3y2 =  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 l1 and l2  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 (l1) is parallel to this line

Slope of l1 is  = m = -2/3

The equation of line (l1) passing through (3, 4) ≡ (x1, y1) is

(y – y1)  = m(x – x1)

∴    (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 (l2) is perpendicular to this line

Slope of l2 is = m = -5/3

The equation of line (l2) passing through  (3, 4) ≡ (x1, y1) is

(y – y1)  = m(x – x1)

∴    (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

∴    10x2 + 6xy – 54x + 15xy + 9y2 – 81y – 90x -54 y + 486= 0

∴     10x2 + 21xy + 9y2 – 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 l1 and l2  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 (l1) is perpendicular to this line

Slope of l1 is  = m = 2

The equation of line (l1) passing through (-1, 2) ≡ (x1, y1) is

(y – y1)  = m(x – x1)

∴     (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 (l2) is perpendicular to this line

Slope of l2 is = m = -4/3

The equation of line (l2) passing through (-1, 2) ≡ (x1, y1) is

∴     (y – y1)  = m(x – x1)

∴     (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

∴     8x2 + 6xy – 4x – 4xy – 3y2 + 2y + 16x + 12 y – 8 = 0

∴      8x2 + 2xy – 3y2 + 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 l1 and l2  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 (l1) is parallel to this line

Slope of l1 is  = m = 1/2

The equation of line (l1) passing through (3, 2) ≡ (x1, y1) is

(y – y1)  = m(x – x1)

∴     (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 (l2) 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

∴     x2 – 2xy + x – 3x + 6y – 3 = 0

∴     x2 – 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 l1 and l2  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 (l1) is perpendicular to this line

Slope of l1 is  = 2/3 = m

The equation of line (l1) passing through (1, 2) ≡ (x1, y1) is

(y – y1)  = m(x – x1)

∴    (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 (l2) is perpendicular to this line

Slope of l2 is = -5/2 = m

The equation of line (l2) passing through (3, 4) = (x1, y1) is

(y – y1)  = m(x – x1)

∴    (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

∴    10x2 + 4xy – 18x – 15xy – 6y2 + 27y + 20x + 8y – 36= 0

∴    10x2 – 11xy – 6y2 + 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 l1 and l2  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 (l1) is perpendicular to this line

Slope of l1 is  = 2/3 = m

The equation of line (l1) passing through (1, 2) = (x1, y1) is

(y – y1)  = m(x – x1)

∴    (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 (l2) is perpendicular to this line

Slope of l2 is – 3 = m

The equation of line (l2) passing through (3, 4) = (x1, y1) is

(y – y1)  = m(x – x1)

∴    (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

∴    6x2 + 2xy – 18x – 9xy – 3y2 + 27y + 15x + 5y – 45= 0

∴    6x2 –  7xy – 3y2 – 3x + 32y – 45= 0

This is the required combined equation.

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