Q. 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 $3x - 4y - 8 = 0$, then
A) $x > \frac{8}{15},\ y < - \frac{8}{5}$
B) $x > \frac{8}{5},\ y < - \frac{8}{15}$
C) $x = \frac{8}{15},\ y = - \frac{8}{5}$
D) None of these
Explanation
(a)
Let $L_{1} \equiv 3x - 4y - 8 = 0$
At point (3,0),
$L_{1} \equiv 9 - 16 - 8 = - 15 < 0$
At point $(x,\ y)$ and (3, 4) opposite sides of $L_{1}$
$\therefore 3x - 4y - 8 > 0\$ …(i)
$\Rightarrow \ 3x - 4( - 3x) - 8 > 0\ \ \lbrack\because y = - 3x\rbrack$
$\Rightarrow \ 15x - 8 > 0\ \ \ \ x > \frac{8}{15}$
Again from Eq. (i),
$3\left( - \frac{y}{3} \right) - 4y - 8 > 0$
$\Rightarrow \ - 5y - 8 > 0\ \Rightarrow \ y < - \frac{8}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.