Question

lines `L_1:ax+by+c=0` and `L_2:lx+my+n=0` intersect at the point `P` and make a angle `theta` between each other. find the equation of a line `L`different from `L_2` which passes through `P` and makes the same angle `theta` with `L_1`

Answer 1

let the equation be `alx+bly+cl=0` (i)and`alx+amy+an=0 (ii)`
on solving simultaneously
`y=(an-cl)/(bl-am)` putting y in (i) we will get, `x=(cm-bm)/(bl-am)` so the point of intersection is
`P=(an-cl)/(bl-am),(cm-bm)/(bl-am)`
`y=y_1=m(m-x_1)` (iii)
since m= `tan theta` so
m1=-a/b m2=-1/m
`tan theta=` `Abs((-b/a+1/m)/(1+al/bm))` now put the value of m in eqn(iii) and put the point P, which is intersection you will get the final answer