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 $L:3x=4y+8$
$=3x-4y-8$
$L_{(1,1)}=3-4-8=-9<0$
Now, $L_{(\alpha ,\beta )}>0$
$3x-4y-8>0$
$\Rightarrow 3x-4(-3x)-8>0[\because y=-3x]$
$15x-8>0$
$\Rightarrow x>\frac{8}{15}$
$\Rightarrow \alpha >\frac{8}{15}$
$3x-4y-8>0$
$\Rightarrow -5y-8>0[\because 3x=-y]$
$\Rightarrow 5y+8<0$
$y<-\frac{8}{5}\Rightarrow \beta <\frac{-8}{5}$