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:
Algorithm:
Example 01:
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:
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:
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:
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:
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:
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:
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:
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:
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:
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:
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:
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. Science > Mathematics > Pair of Straight Lines > Combined Equation of Lines When Point and Two Other Lines are Given |