Let for any three distinct consecutive terms a, b, c of an A.P, the lines a x+b y+c=0 be c…

Question

Let for any three distinct consecutive terms $a, b, c$ of an A.P, the lines $a x+b y+c=0$ be concurrent at the point $P$ and $$Q(\alpha, \beta)$$

Accepted Answer

$\because a,b,c$ and in A.P

$\Rightarrow 2b=a+c\Rightarrow a-2b+c=0$

$	herefore ax+by+c$ passes through fixed point $(1,-2)$

$	herefore P=(1,-2)$

For infinite solution,

$\begin{gathered} D={D}_{1}={D}_{2}={D}_{3}=0 \ D:|\begin{array}{ccc}1 \quad  1 \quad  1 \ 2 \quad  5 \quad  \alpha \ 1 \quad  2 \quad  3\end{array}|=0 \ \Rightarrow \alpha =8 \ {D}_{1}:|\begin{array}{ccc}6 \quad  1 \quad  1 \ \beta  \quad  5 \quad  \alpha \ 4 \quad  2 \quad  3\end{array}|=0\Rightarrow \beta =6 \ 	herefore Q=(8,6) \ 	herefore {Q}^{2}=113 \end{gathered}$

Solutions · 1