Question

If l sin theta + m cos theta +n =0 and l'sin theta + m' cos theta + n' = 0,then prove that (mn'- m'n)^2 + (nl'- ln')^2 = (lm'- l'm)^2

Answer 1

l sinθ + m cosθ + n = 0
l' sinθ + m' cosθ + n' = 0

Let x = sinθ and y = cosθ
then, equation will be
lx + my + n = 0
l'x + m'y + n' = 0
now use Cramer rule to solve it.
x/(mn' - m'n) = -y/(n'l - nl') = 1/(lm' - l'm)
then, x = (mn' - m'n)/(lm' - l'm)
y = -(n'l - nl')/(lm' - l'm) = (nl' - n'l)/(lm' - l'm)

because x = sinθ and y = cosθ
and we know, sin²θ + cos²θ = 1
so, x² + y² = 1
⇒{(mn' - m'n)/(lm' - l'm)}² + {(nl' - n'l)/(lm' - l'm)}² = 1
⇒(mn' - m'n)² + (nl' - n'l)² = (lm' - l'm)²

hence, proved that ,
(mn' - m'n)² + (nl' - n'l)² = (lm' - l'm)²