Let R be the interior region between the lines 3x-y+1=0 and x+2y-5=0 containing the origin…

Question

Let $R$ be the interior region between the lines $3x-y+1=0$ and $x+2y-5=0$ containing the origin. The set of all values of $a$, for which the points $({a}^{2},a+1)$ lie in $R$, is :

Accepted Answer

$\begin{gathered} P({a}^{2},a+1) \ {L}_{1}=3x-y+1=0 \end{gathered}$

Origin and $P$ lies same side w.r.t. ${L}_{1}$

$\begin{gathered} \Rightarrow {L}_{1}(0)\cdot {L}_{1}(P)>0 \ \therefore 3({a}^{2})-(a+1)+1>0 \end{gathered}$

$\begin{gathered} \Rightarrow 3{a}^{2}-a>0 \ a\in (-\infty,0)\cup (\frac{1}{3},\infty)\, \end{gathered}$

Let ${L}_{2}:x+2y-5=0$

Origin and $P$ lies same side w.r.t. ${L}_{2}$

$ $$\& \Rightarrow {L}_{2}(0)\cdot {L}_{2}(P)>0  \  \& \Rightarrow {a}^{2}+2(a+1)-5$\therefore a\in (-3,1)\,$$$

Intersection of (1) and (2)

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Solutions · 1