Pair of Straight LinesThe equation ax^{2}+ 2hxy + by^{2}+ 2gx + 2fy + c = 0. Represents a second degree equation where a, h, b doesn’t variables simultaneously.Let a ≠ 0.Now, the above equation becomesa^{2}x^{2}+ 2ax (hy + g) = aby^{2}– 2afy – acon completing the square on the left side, we get,a^{2}x^{2}+ 2ax (hy + g) = y^{2}(h^{2}– ab) + 2y (gh – af) + g^{2}– ac.i.e. (ax + hy + g) = + √y^{2}(h^{2}–ab)+2y(gh–af)g^{2}–acWe cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign be a perfect square. The condition for this is,(gh – af)^{2}= (h^{2}– ab) (g^{2}– ac)i.e. g^{2}h^{2}– 2afgh + a^{2}f^{2}= g^{2}h^{2}– abg^{2}– abg^{2}– ach^{2}+ a2bccancelling and diving by a, we have the required conditionabc + 2fgh – af^{2}– af^{2}– bg^{2}– ch^{2}= 0Illustration:What is the point of intersection of two straight lines given by general equation ax^{2}+ 2hxy + by^{2}+ 2gx + 2fy + c = 0?Solution:The general solution isax^{2}+ 2hxy + by^{2}+ 2gx + 2fy + c = 0 …… (1)Let (α, ß) be the point of intersection we consider line paralleled transformation.x = x’ + α, y = y’ + ßFrom (1) we havea(x’ + α)^{2}+ 2h(x’ + α) (y’ + ß) + b(y’ + ß)^{2}+ 2g(x’ + α) + 2f(y’ + ß) + c = 0⇒ ax’^{2}+ 2hx’y’ + by’^{2}+ a α^{2}+ 2hαß + bß^{2}+ 2gα + 2fß + 2x’(a α + hß + g) + 2y’ + 2y’ (hα + bß + f) = 0⇒ ax’^{2}+ 2hx’y’ + by’^{2}+ 2x’(aα + hß + g) + 2g’ + 2y’ (hα + bß + f) = 0Which must be in the formax'^{2}+ 2hx’y’ + by’ = 0This cannot be possible unlessaα + hß + g = 0hα + bß + f = 0Solvingα/hf–bg = ß/hg–af = 1/ab–h^{2}α = hf–bg/ab–b^{2}, ß = hg–af/ab–h^{2}Illustration:Represent lines y = 2x and y = 3x by a homogeneous equation of second degreeSolution:(y – 2x) (y – 3x) = 0Or 6x^{2}– 5xy + y^{2}= 0Illustration:Represent lines parallel to y = 2x and y = 3x by a second degree equationSolution:(y – 2x – c_{1}) (y – 3x – c_{2}) (where c_{1}and c_{2}are constants)= 6x^{2}– 5xy + y^{2}+ (3c_{1}+ 2c_{2}) x + (– c_{1}– c_{2}) y + c_{1}c_{2}= 0Note:1. Homogeneous part is same as for the equation of above illustration. Therefore, the homogeneous part of a general second degree equation determines the slope of the lines i.e. lines parallel to ax^{2}+ 2hxy + by^{2}+ c = 0 and through the origin are represented by the equation ax^{2}+ 2hxy + by^{2}= 02. The equation ax^{2}+ 2hxy + by^{2}+ 2fy + c = 0 represents a pair of parallel straight lines if h/a = b/h = f/g or bg^{2}= af^{2}The distance between them is given by 2√g^{2}–ac/a(a+b) or √f^{2}–bc/b(a+b)Illustration:Does the second degree equation x^{2}+ 3xy + 2y^{2}– x – 4y – 6 = 0 represents a pair of lines. If yes, find their point of intersection.Solution:We observe thata = 1, h = 3/2, b = 2, g = –1/2, f = 2, c = – 6\ abc + 2fgh – af^{2}– bg^{2}– ch^{2}= – 12 + 3 – 4 – 1/2 + 27/2 = 0Therefore the given second-degree equation represents a pair of lines, x^{2}+ 3xy + 2y^{2}– x – 4y – 6 = (x + 2y + 2) (x + y – 3).Consider the equations formed by first two rows of .i.e. ax + hy + g = 0 and hx + by + f = 0i.e. x + 3/2 y – 1/2 and 3/2 x + 2y – 2 = 0Solving these, we get the required point of intersection.i.e. 2x + 3y – 1 = 03x – 4y – 4 = 0Solving the above equation, we get x = 8, y = –5.Note:(2x + 3y – 1)(3x + 4y – 4) ≠ x

^{2}+ 3xy + 2y^{2}– x – 4y – 6.