Straight Lines Question 3
Question 3 - 2024 (27 Jan Shift 2)
Let $\mathrm{R}$ 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 $\mathrm{P}$ lies same side w.r.t. $\mathrm{L}_{1}$
$\Rightarrow \mathrm{L}{1}(0) \cdot \mathrm{L}{1}(\mathrm{P})>0$
$\therefore 3\left(\mathrm{a}^{2}\right)-(\mathrm{a}+1)+1>0$
$\Rightarrow 3 \mathrm{a}^{2}-\mathrm{a}>0$
$\mathrm{a} \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \ldots$
Let $\mathrm{L}_{2}: \mathrm{x}+2 \mathrm{y}-5=0$
Origin and $\mathrm{P}$ lies same side w.r.t. $\mathrm{L}_{2}$
$$ \begin{align*} & \Rightarrow \mathrm{L}{2}(0) \cdot \mathrm{L}{2}(\mathrm{P})>0 \ & \Rightarrow \mathrm{a}^{2}+2(\mathrm{a}+1)-5<0 \ & \Rightarrow \mathrm{a}^{2}+2 \mathrm{a}-3<0 \ & \Rightarrow(\mathrm{a}+3)(\mathrm{a}-1)<0 \ & \therefore \mathrm{a} \in(-3,1) \ldots \ldots \ldots \ldots \tag{2} \end{align*} $$
Intersection of (1) and (2)
$\mathrm{a} \in(-3,0) \cup\left(\frac{1}{3}, 1\right)$
