Here, you will learn equation of family of lines and the combined equation of the bisectors of the angle between the lines.
Family of Lines Equation
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.
Example : 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.
Solution : The given family of straight lines can be rewritten as
(3x+2y+1)\(sin\theta\) + (4x-7y+2)\(cos\theta\) = 0
or, (4x-7y+2) + \(tan\theta\)(3x+2y+1) = 0 which is of the form \(L_1\) + \(\lambda L_2\) = 0
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
i.e. (\(-11\over 29\),\(2\over 29\))
Equation of bisectors of angles between two lines :
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 :
\(a_1x+b_1y+c_1\over {\sqrt{{a_1}^2 + {b_1}^2}}\) = \(\pm\) \(a_2x+b_2y+c_2\over {\sqrt{{a_2}^2 + {b_2}^2}}\) …..(i)
(a) Equation of bisector of angle containing origin : 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)
(b) Equation of bisector of acute/obtuse angle : To find the equation of the bisector of the acute or obtuse angle :
(i) Let \(\phi\) be angle between one of the two bisectors and one of the two given lines. Then if tan\(\phi\) < 1
i.e. \(\phi\) < 45 i.e. 2\(\phi\) < 90, the angle bisector will be bisector of acute angle.
(ii) 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.
Determine the sign of \(a_1a_2+b_1b_2\)
| If sign of \(a_1a_2+b_1b_2\) | for obtuse angle bisector | for acute angle bisector |
|---|---|---|
| + | use + sign in eq.(1) | use – sign in eq.(1) |
| – | use – sign in eq.(1) | use + sign in eq.(1) |
The combined equation of angle bisectors :
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.
Note :
(i) If a = b, then bisectors are \(x^2-y^2\) = 0 i.e. x – y = 0, x + y = 0
(ii) If h = 0, the bisectors are xy = 0 i.e. x = 0, y = 0
(iii) The two bisectors are always at right angles, since we have coefficient of \(x^2\) + coefficient of \(y^2\) = 0