Question

Let m be any real number. Then the equation (m+1)^2 x-my-m^2-1=0 is a line. There is a unique point which this line goes through no matter what the value of m is; find the coordinates of this point.

Answer 1

If true then we should be able to pick any value of m
let m=0 ----> x -1 = 0 or x = 1
let m=-1 ---> y - 1 -1 = 0 or y = 2
the point is (1,2)

Proof:
sub (1,2) into original equation
LS = (1)(m+1)^2 - 2m - m^2 - 1
= m^2 + 2m + 1 - 2m - m^2 - 1
= 0
= RS