Question

If $P (\alpha, \beta)$ be a point on the line $3x+y = 0$ such that the point $P$ and the point $Q (1.1)$ lie on either side of the line $3x = 4y + 8$. then

• A. $\alpha > \frac{8}{15} , \beta < \frac{-8}{5} $
• B. $\alpha < \frac{8}{15} , \beta < \frac{-8}{5} $
• C. $\alpha > \frac{8}{15} , \beta > \frac{-8}{5} $
• D. $\alpha < \frac{8}{15} , \beta > \frac{-8}{5} $

Answer 1

Answer: (A)

Line $L: 3 x=4 y+8$
$=3 x-4 y-8$
$ L_{(1,1)}=3-4-8=-9<0$
Now, $L_{(\alpha, \beta)}>0$
$3 x-4 y-8>0$
$\Rightarrow 3 x-4(-3 x)-8 > 0 [\because y=-3 x]$
$ 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 [ \because 3x = -y]$
$ \Rightarrow 5 y+8 < 0 $
$ y < - \frac{8}{5} \Rightarrow \beta < \frac{-8}{5}$