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)

$x>\frac{8}{15}$
$y<\frac{-8}{5}$

Let $L=3x-4y-8=0$ be the equation of line and $P(x,y)$ and $Q(3,4)$ are given points.
Then $P$ and $Q$ are on opposite side of the line iff $\frac{L_{11}}{L_{22}}<0$
$P$ is a point on the line $y=-3x.$
i.e $P=(x,-3x)$
$\therefore \frac{L_{11}}{L_{22}}=\frac{15x-8}{-15}<0$
$\Rightarrow x>\frac{8}{15}$
$\Rightarrow y=-3x<\frac{-8}{5}$
Hence, options A and C.