Family of lines:
Assume that L1= a1x+b1y+c1 = 0 and L2=a2x + b2y + c2 = 0 are the two intersecting lines and let the point of their intersection be (α, β). Now if we write these 2 equations in this form L1 + λL2 (where λ is a parameter)
.........(1)
![666_Family of lines1.png](https://www.expertsmind.com/../CMSImages/666_Family of lines1.png)
then for the different values of λ, (1) will give different straight lines.
Now ![122_Family of lines2.png](https://www.expertsmind.com/../CMSImages/122_Family of lines2.png)
![1344_Family of lines3.png](https://www.expertsmind.com/../CMSImages/1344_Family of lines3.png)
=>(α, β) always lies on (1) does not matter what is the value of λ.
Therefor (1) represent a family of straight lines passing through the point of intersection of a1x+b1y+c1 = 0 and a2x + b2y + c2 = 0.
Note:
- When we have to show that the line always passes through a fixed point, we use the concept of family of lines.
- Family of lines perpendicular to the given line ax+by+c = 0 is given by bx-ay+k=0, here k is a parameter.
- Family of lines parallel to the given line ax+by+c = 0 can be given by ax+by+k=0, where k is a parameter.
Illustration: Find the equation of straight line which belongs to both the family of lines 5x + 3y - 2 + l1 (3x - y - 4) = 0 and x - y + 1 + l2 (2x - y - 2) = 0
Solution. Lines of 1st family are concurrent at (1, -1) and that of 2nd at (3, 4)
∴ The required line passes through both these points
∴ Equation is 5x - 2y - 7 = 0
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