If $P(\alpha, \beta)$ be a point on the line $3 x+y=0$ such that the point $P$ and the point $Q(1,1)$ lie on…
If $P(\alpha, \beta)$ be a point on the line $3 x+y=0$ such that the point $P$ and the point $Q(1,1)$ lie on either side of the line $3 x=4 y+8$, then
- $\alpha>\frac{8}{15}, \beta < \frac{-8}{5}$
- $\alpha < \frac{8}{15}, \beta < \frac{-8}{5}$
- $\alpha>\frac{8}{15}, \beta>\frac{-8}{5}$
- $\alpha < \frac{8}{15}, \beta>\frac{-8}{5}$
Solution
Line $L: 3 x=4 y+8$
$
\begin{aligned}
& =3 x-4 y-8 \\
L_{(1,1)} & =3-4-8=-9 < 0
\end{aligned}
$
Now, $\quad L_{(\alpha, \beta)}>0$
$
\begin{aligned}
& 3 x-4 y-8>0 \\
\Rightarrow & 3 x-4(-3 x)-8>0 \quad[\because y=-3 x]
\end{aligned}
$
$\begin{array}{rlrl} & 15 x-8>0 \Rightarrow x>\frac{8}{15} \Rightarrow \alpha>\frac{8}{15} & \\ & 3 x-4 y-8 & >0 \\ \Rightarrow & -5 y-8 & >0 \\ 5 y+8 & < 0 \\ y & < -\frac{8}{5} \Rightarrow \beta < \frac{-8}{5}\end{array} \quad[\because 3 x=-y]$
Asked in: AP EAMCET 2019 (21 Apr Shift 1)
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