Science > Mathematics > Pair of Straight Lines > Combined Equation of Lines When Point and Two Other Lines are Given
In this article, we shall study to find a combined or joint equation of pair of lines when a point on the line and equations of two other lines are given.
You Should Know It:
- 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 the slope of one line is m, then the slope of a line perpendicular to it is – 1/m.
Algorithm:
- Find the slopes of the given line using the formula given in point 1 of notes.
- Find the slopes of required lines forming a pair using relations given in points 2 and 3 of notes.
- 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.
- 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 01:
- Find the joint equation of a pair of lines through the origin such that one is parallel to x + 2y = 5 and the 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 the 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 the slope of this line is -2/-1 = 2
As required first line (l2) is perpendicular to this line
The 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 02:
- 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 03:
- Find the joint equation of a pair of lines through the origin such that one is parallel to 3x – y = 7 and the 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 04:
- 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 05:
- 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 06:
Example 35:
- Find the joint equation of a pair of lines through the origin such that one is parallel to and the 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 07:
- 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 08:
- 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 the 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 09:
- 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 10:
- Find the joint equation of a pair of lines through the point (3, 2) one of which is parallel to x – 2y = 2 and the 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 11:
- 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 12:
- 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.