{"id":3902,"date":"2021-08-10T01:25:13","date_gmt":"2021-08-10T01:25:13","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3902"},"modified":"2021-11-30T16:36:26","modified_gmt":"2021-11-30T11:06:26","slug":"family-of-lines","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/family-of-lines\/","title":{"rendered":"Family of Lines – The combined Equation of Angle Bisectors"},"content":{"rendered":"
Here, you will learn equation of family of lines and the combined equation of the bisectors of the angle between the lines.<\/p>\n
If equation of two lines be P = \\(a_1x+b_1y+c_1\\) = 0 and Q = \\(a_2x+b_2y+c_2\\) = 0, then the equation of the lines passing through the point of intersection of these lines is : P + \\(\\lambda\\)Q = 0 or \\(a_1x+b_1y+c_1\\) + \\(\\lambda\\)\\(a_2x+b_2y+c_2\\) = 0. The value of \\(\\lambda\\) is obtained with the help of the additional information given in the problem.<\/p>\n\n\n
Example : <\/span> Prove that each member of the family of straight lines (\\(3sin\\theta + 4cos\\theta\\))x + (\\(2sin\\theta – 7cos\\theta\\))y + (\\(sin\\theta + 2cos\\theta\\)) = 0 (\\(\\theta\\) is a parameter) passes through a fixed point.<\/p>\n Solution : <\/span>The given family of straight lines can be rewritten as If equation of two intersecting lines are \\(a_1x+b_1y+c_1\\) = 0 and \\(a_2x+b_2y+c_2\\) = 0, then equation of bisectors of the angles between these lines are written as :<\/p>\n \\(a_1x+b_1y+c_1\\over {\\sqrt{{a_1}^2 + {b_1}^2}}\\) = \\(\\pm\\) \\(a_2x+b_2y+c_2\\over {\\sqrt{{a_2}^2 + {b_2}^2}}\\)\u00a0 \u00a0…..(i)<\/p>\n (a)\u00a0 Equation of bisector of angle containing origin :\u00a0 If the equation of the lines are written with constant terms \\(c_1\\) and \\(c_2\\) positive, then the equation of the bisectors of the angle containing the origin is obtained by taking positive sign in (i)<\/p>\n (b)\u00a0 Equation of bisector of acute\/obtuse angle : To find the equation of the bisector of the acute or obtuse angle :<\/p>\n (i)\u00a0 Let \\(\\phi\\) be angle between one of the two bisectors and one of the two given lines. Then if tan\\(\\phi\\) < 1<\/p>\n i.e. \\(\\phi\\) < 45 i.e. 2\\(\\phi\\) < 90, the angle bisector will be bisector of acute angle.<\/p>\n (ii)\u00a0 See whether the constant terms \\(c_1\\) and \\(c_2\\) in the two equation are +ve or not. If not then multiply both sides of given equation by -1 to make the constant terms positive.<\/p>\n Determine the sign of\u00a0 \\(a_1a_2+b_1b_2\\)<\/p>\n\n\n The combined equation of angle bisectors between the lines represented by homogeneous equation of second degree is given by \\(x^2-y^2\\over {a-b}\\) = \\(xy\\over h\\), a \\(\\ne\\) b, h \\(\\ne\\) 0.<\/p>\n Note :<\/strong><\/p>\n (i)\u00a0 If a = b, then bisectors are \\(x^2-y^2\\) = 0 i.e. x – y = 0, x + y = 0<\/p>\n (ii)\u00a0 If h = 0, the bisectors are xy = 0 i.e. x = 0, y = 0<\/p>\n (iii)\u00a0 The two bisectors are always at right angles, since we have coefficient of \\(x^2\\) + coefficient of \\(y^2\\) = 0<\/p>\n\n\n
\n (3x+2y+1)\\(sin\\theta\\) + (4x-7y+2)\\(cos\\theta\\) = 0
\n or, (4x-7y+2) + \\(tan\\theta\\)(3x+2y+1) = 0 which is of the form \\(L_1\\) + \\(\\lambda L_2\\) = 0
\n Hence each member of it will pass through a fixed point which is the intersection of 4x-7y+2 = 0 and 3x+2y+1 = 0
\n i.e. (\\(-11\\over 29\\),\\(2\\over 29\\))
<\/p>\n\n\nEquation of bisectors of angles between two lines :<\/h2>\n
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\n If sign of \\(a_1a_2+b_1b_2\\)<\/th>\n for obtuse angle bisector<\/th>\n for acute angle bisector<\/th>\n <\/tr>\n \n +<\/td>\n use + sign in eq.(1)<\/td>\n use – sign in eq.(1)<\/td>\n <\/tr>\n \n –<\/td>\n use – sign in eq.(1)<\/td>\n use + sign in eq.(1)<\/td>\n <\/tr>\n <\/tbody><\/table>
\n\n\nThe combined equation of angle bisectors :<\/h2>\n