Suppose a, b and c are distinct real numbers. Let Δ=aa+ca−bb−cba+bc+bc−ac=0 .Then the straight line a(x−5)+b(y−2)+c=0 passes through the fixed point
(5, 2)
(6, 2)
(6, 3)
(5, 3)
Applying C1→aC1+bC2+cC3 we get
Let Δ=1aa(a+b+c)a+ca−bb(a+b+c)ba+bc(a+b+c)c−ac
=1aa(a+b+c)Δ1, where
Δ1=aa+ca−bbba+bcc−ac
Using C2→C2−C1,C3→C3−C1 we get
Δ1=ac−bb0ac−a0=aa2+b2+c2∴ Δ=(a+b+c)a2+b2+c2=0
As a, b, c are distinct real numbers, a2+b2+c2≠0 therefore
a+b+c=0
⇒the line a(x–5)+b(y–2)+c=0 passes through (6, 3)