Question

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

• A. $x>\frac{8}{5},y<\frac{-8}{15}$
• B. $x>\frac{8}{15},y<-\frac{8}{5}$
• C. $x=\frac{8}{15},y=\frac{-8}{5}$
• D. $x=2,y=-2$

Answer 1

The correct option is B $x>\frac{8}{15},y<-\frac{8}{5}$

Let $k=3x-4y-8$

then the value of $k$ at $(3,4)=3\times 3-4\times 4-8=-15<0$

$\therefore$ For the point $P(x,y)$ we should have $k>0$

$\Rightarrow 3x-4y-8>0$

$\Rightarrow 3x-4(-3x)-8>0$ $[\because P(x,y)$ lies on $y=-3x]$

$\Rightarrow x>\frac{8}{15}$

and $\Rightarrow -y-4y-8>0\Rightarrow y<-\frac{8}{5}.$