Straight Lines Question 3
Question 3 - 2024 (27 Jan Shift 2)
Let $R$ be the interior region between the lines $3 x-y+1=0$ and $x+2 y-5=0$ containing the origin. The set of all values of $a$, for which the points $\left(a^{2}, a+1\right)$ lie in $R$, is :
(1) $(-3,-1) \cup\left(-\frac{1}{3}, 1\right)$
(2) $(-3,0) \cup\left(\frac{1}{3}, 1\right)$
(3) $(-3,0) \cup\left(\frac{2}{3}, 1\right)$
(4) $(-3,-1) \cup\left(\frac{1}{3}, 1\right)$
Show Answer
Answer (2)
Solution
$P\left(a^{2}, a+1\right)$
$L _1=3 x-y+1=0$
Origin and $P$ lies same side w.r.t. $L _1$
$\Rightarrow L _1(0) \cdot L _1(P)>0$
$\therefore 3\left(a^{2}\right)-(a+1)+1>0$
$\Rightarrow 3 a^{2}-a>0$
$a \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \ldots$
Let $L _2: x+2 y-5=0$
Origin and $P$ lies same side w.r.t. $L _2$
$$ \begin{align*} & \Rightarrow L _2(0) \cdot L _2(P)>0 \\ & \Rightarrow a^{2}+2(a+1)-5<0 \\ & \Rightarrow a^{2}+2 a-3<0 \\ & \Rightarrow(a+3)(a-1)<0 \\ & \therefore a \in(-3,1) \ldots \ldots \ldots \ldots \tag{2} \end{align*} $$
Intersection of (1) and (2)
$a \in(-3,0) \cup\left(\frac{1}{3}, 1\right)$
