S. Agarwal
Last Activity: 10 Years ago
Find the equation of the line which passes through the point of intersection of L1:2x – 3y + 5 = 0 and L2: 3x – 2y + 5 = 0, and also passes through the origin.
Here’s one way to solve the problem.
(i) Find the point of intersection L1 and L2: This point will be (-1, 1)
(ii) Use the two point form to find the required equation: (y – 0)=[(1-0)/(-1-0)](x – 0) or x + y = 0.
But that’s a boring method. I’m interested in family of lines. So here’s another method.
As discussed before, equation of any line passing through the point of intersection of the lines L1=0 and L2 =0 will be of the form L1+λL2 =0, where λ is a parameter.
That is, the required line will be of the form (2x – 3y + 5)+λ(3x – 2y + 5) = 0. That’s half the work done! But what about λ?
Another condition is given to us: The line passes through the origin as well. We can therefore substitute the coordinates to obtain the value of λ.
We have,
2(0)-3(0)+5 + λ(3(0)-2(0)+5) = 0,
or λ=-1.
And we’re done!
The required line is (2x – 3y + 5)+(-1)(3x – 2y + 5) = 0, or x + y = 0.
We didn’t even have to find the point of intersection of the given lines. Saved some effort.
So remember, any line passing through the point of intersection of two given lines will be given by L1+λL2 =0, and the value of the parameter λ will be determined by another given condition