Question

If $P$ is a point $(x,y)$ on the line $y=-3x$ such that $P$ and the point $(3,4)$ are on the opposite sides of the line $3x-4y=8$, then

• A. $x>\frac{8}{15}$
• B. $x>\frac{8}{5}$
• C. $y<-\frac{8}{5}$
• D. $y<-\frac{8}{15}$

Answer 1

Answer: (A), (C)

The correct options are
A $x>\frac{8}{15}$
C $y<-\frac{8}{5}$

Let $L=3x-4y-8$. Then the value of $L$ at $(3,4)$ is $3\times 3-4\times 4-8=-15<0$. Hence, for the point $P(x,y)$, we should have
$L>0$
or $3x-4y-8>0$
or $3x-4(-3x)-8>0$
($\because P(x,y)$ lies on $y=-3x$)
$\Rightarrow x>\frac{8}{15}$
and $-y-4y-8>0$
$\Rightarrow y<-\frac{8}{5}$