If a point $(\alpha, \beta)$ of $3 x+y=0$ and $(3,4)$ lie on the opposite sides of $3 x-4 y-8=0$ then which…
If a point $(\alpha, \beta)$ of $3 x+y=0$ and $(3,4)$ lie on the opposite sides of $3 x-4 y-8=0$ then which of the following is correct?
- $15 \alpha-8>0$
- $\alpha \in(-\infty, \infty)$
- $15 \alpha-8=0$
- $\alpha=0$
Solution
Given line $3 \mathrm{x}+\mathrm{y}=0$ with $(\alpha, \beta)$
and line $3 x-4 y-8=0$
Satisty $(3,4)$ in the eq $3 x-4 y-8=0$
$
\begin{aligned}
& \mathrm{L}_2=A_{-}(\varepsilon) \times \varepsilon_{-}(\Gamma) \times r \\
& \Rightarrow \quad 9-16-8 \\
& \Rightarrow \quad-7-8=-15 < 0 \\
&
\end{aligned}
$
Now, satisty the line $3 x+y=0$ on the equation
$
\begin{aligned}
& 3 x-4 y-8=0 . \\
& \text { Here, } L(\alpha, \beta)>0 \\
& 3 x-4(-3 x)-8>0 \\
& 3 x+12 x>8 \\
& 15 x>8
\end{aligned}
$
Satisty $\mathrm{x}=\alpha$
Then, $15 \alpha>8 \Rightarrow 15 \alpha-8>0$
Asked in: AP EAMCET 2022 (06 Jul Shift 1)
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